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Electrostatic Potential and Capacitance

Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid-point.

(a) Potential at the mid-point of the line joining the two charges is

box enclose straight V space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction open square brackets straight q subscript 1 over straight r subscript 1 plus straight q subscript 2 over straight r subscript 2 close square brackets end enclose space space space space space space space space space space space space space space space space space space equals space 9 space cross times space 10 to the power of 9 space open square brackets fraction numerator 15 space cross times space 10 to the power of negative 6 end exponent over denominator 0.15 end fraction plus fraction numerator 2.5 space cross times space 10 to the power of negative 6 end exponent over denominator 0.15 end fraction close square brackets straight V space space space space space space space space space space space space space space space space space space space equals space 9 space cross times space 10 to the power of 9 space cross times space 10 to the power of negative 6 end exponent open square brackets 10 plus 50 over 3 close square brackets space space space space space space space space space space space space space space space space space space space space equals space 9 space cross times space 10 cubed cross times 80 over 3

or,

straight V space equals space 2.4 space cross times space 10 to the power of 5 straight V

(a) Potential at the mid-point of the line joining the two charges is

space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q subscript 1 over straight r subscript 1 squared equals space 9 space cross times space 10 to the power of 9 space cross times space fraction numerator 1.5 space cross times 10 to the power of negative 6 end exponent over denominator left parenthesis 0.15 right parenthesis squared end fraction equals space space 6 space cross times space 10 to the power of 5 space Vm to the power of negative 1 end exponent space along space OB

Electric field at the mid-point O due to charge at B

space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q subscript 2 over straight r subscript 2 squared space equals space 9 space cross times space 10 to the power of 9 space cross times space fraction numerator 2.5 space cross times space 10 to the power of negative 6 end exponent over denominator left parenthesis 0.15 right parenthesis squared end fraction space equals space 10 space cross times space 10 to the power of 5 space Vm to the power of negative 1 end exponent space along space OA

Thus, the total electric field at the mid-point O is

straight E space equals space 10 space cross times space 10 to the power of 5 space minus space 6 space cross times space 10 to the power of 5 space space space space equals space 4 space cross times space 10 to the power of 5 Vm to the power of negative 1 end exponent left parenthesis along space BA right parenthesis

(b) Potential at the point C due to the two charges is

straight V space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction open square brackets straight q subscript 1 over straight r subscript 1 plus straight q subscript 2 over straight r subscript 2 close square brackets space equals space 9 cross times 10 to the power of 9 open square brackets fraction numerator 1.5 space cross times space 10 to the power of negative 6 end exponent over denominator square root of 325 cross times 10 to the power of negative 2 end exponent end fraction plus fraction numerator 2.5 space cross times space 10 to the power of negative 6 end exponent over denominator square root of 325 cross times 10 to the power of negative 2 end exponent end fraction close square brackets straight V equals space fraction numerator 9 cross times 10 to the power of 9 cross times 10 to the power of negative 6 end exponent over denominator 10 to the power of negative 2 end exponent end fraction. space fraction numerator 4.0 over denominator square root of 325 end fraction straight V equals space fraction numerator 9 space cross times space 4 over denominator 18.02 end fraction space cross times space 10 to the power of 5 straight V space equals space 2 space cross times space 10 to the power of 5 space straight V

Electric field at C due to charge at A
(a) Potential at the mid-point of the line joining the two charges is

Electric field at C due to charge at B

straight E subscript 2 space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q subscript 2 over straight r subscript 2 squared space equals space 9 space cross times space 10 to the power of 9 cross times fraction numerator 2.5 space cross times space 10 to the power of negative 6 end exponent over denominator 325 space cross times space 10 to the power of negative 4 end exponent end fraction space equals space 6.92 space cross times space 10 to the power of 5 space Vm to the power of negative 1 end exponent

If the angle between E₁ and E₂ be θ, then

tan straight theta over 2 space equals space fraction numerator 0.15 over denominator 0.10 end fraction space equals space 1.5

straight theta divided by 2 space equals space 56.3 space degree space space space or comma space space space space straight theta space equals space 112.6 degree

Thus, magnitude of resultant field at C is
(a) Potential at the mid-point of the line joining the two charges is

Let the field E makes angle α with the field E₁.
(a) Potential at the mid-point of the line joining the two charges is

If field £ makes angle (3 with the direction BA, then
(a) Potential at the mid-point of the line joining the two charges is

Therefore, angle of 69.4° is made by the electric field with the line joining the two charges. q₂ to q₁.

396 Views

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

$\Rightarrow$ $Q = 12 \times 10^{- 8} C$

Electrostatic energy stored in capacitor

U_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 600 \times 10^{- 12} \times (200)^{2} U_{1} = 12 \times 10^{- 6} J

Now, the supply is disconnected and the capacitor is connected to another similar uncharged 600 pF capacitor.
Therefore, the charge is divided equally between the two capacitors.
Hence,

Q_{1} = Q_{2} = \frac{12 \times 10^{- 8}}{2} = 6 \times 10^{- 8} C

and

V_{1} = V_{2} = \frac{Q_{1}}{C_{1}} = \frac{6 \times 10^{- 8}}{600 \times 10^{- 12}} = 100 V

Total capacitance, C = C_{1} + C_{2}

= 600 \times 10^{- 12} F + 600 \times 10^{- 12} F = 1200 \times 10^{- 12} F

Now, the electrostatic energy stored is given as

U_{2} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 1200 \times 10^{- 12} \times (100)^{2}

\Rightarrow

U_{2} = 6 \times 10^{- 6} J

Electrostatic energy lost in the process

= U_{1} - U_{2} = 12 \times 10^{- 6} - 6 \times 10^{- 6} = 6 \times 10^{- 6} J .

571 Views

A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q.
i) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
ii) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

As a ‘+ q’ charge place at the centre of the shell, it will create a ‘–q’ charge on the inner surface of the shell.
The charge on the outer surface will increase by + q due to the –q charge on the inner surface by induction. Therefore, there will be total (Q + q) charge on the outer surface of the shell and –q charge on the inner surface of the shell.
As a ‘+ q’ charge place at the centre of the shell, it will creat
Now surface area of inner surface $space equals space 4 πr subscript 1 squared$
and surface area of outer surface $space equals space 4 πr subscript 2 squared$
Thus, charge density on the outer surface $space equals space fraction numerator straight Q plus straight q over denominator 4 πr subscript 2 squared end fraction$
and charge density on the inner surface $space equals space fraction numerator negative straight q over denominator 4 πr subscript 1 squared end fraction.$

1076 Views

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to these charges array at the centre of the cube.

Diagonal DF of cube

DF space equals space square root of straight b squared plus straight b squared plus straight b squared end root DF space equals space straight b square root of 3

Thus,

DO space equals space DF over 2 space equals space fraction numerator square root of 3 over denominator 2 end fraction straight b

Due to one charge q the potential at the centre O is given by

box enclose straight V space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q over straight r end enclose space equals space open parentheses fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q over denominator begin display style fraction numerator square root of 3 straight b over denominator 2 end fraction end style end fraction close parentheses

Due to eight charges the total potential at the centre O is given as

straight V space equals space 8 open parentheses fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q over denominator begin display style fraction numerator square root of 3 over denominator 2 end fraction end style straight b end fraction close parentheses

space equals space fraction numerator 4 straight q over denominator square root of 3 πε subscript 0 straight b end fraction

Remark: Due to two opposite corners D and F electric field intensity at the centre ‘O’ are equal in magnitude and opposite in direction. Therefore, they cancel out each other. Similarly all other intensities cancel out each other and the total electric field at centre is zero.

1367 Views

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of –2 x 10^–9 C from a point P(0, 0, 3 cm) to a point Q(0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Given charge q = 8 mC = 8 x 10^–1 C is located at origin and the small charge (q₀ = –2 x 10^–9 C) is taken from point P (0, 0, 3 cm) to a point Q (0, 4, cm, 0) through point R (0, 6 cm, 9 cm) which is shown in the figure.
Initial separation between q₀ and q is r_p = 3 cm = 0.03 m
Final separation between q₀ and q is r_Q = 4 cm = 0.4 m
Given charge q = 8 mC = 8 x 10–1 C is located at origin and the sm
Work done in taking the charge q₀ from point P to Q does not dependent on the path followed and depends only upon r_p and r_Q i.e., initial and final positions.
$straight W space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction qq subscript 0 open parentheses 1 over straight r subscript straight Q minus 1 over straight r subscript straight p close parentheses$
or, $straight W space equals space 9 space cross times space 10 to the power of 9 space cross times space 8 space cross times space 10 to the power of negative 3 end exponent space cross times space left parenthesis negative 2 cross times 10 to the power of negative 9 end exponent right parenthesis space cross times space open parentheses fraction numerator 1 over denominator 0.04 end fraction minus fraction numerator 1 over denominator 0.03 end fraction close parentheses space space space space space equals space 1.2 space straight J.$