Question 39: Electrostats & Capacitance

Question

A charge +q is placed at the origin O of XY-axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is

Accepted Answer

Potential at point A,
$${V_A} = {1 \over {4\pi {\varepsilon _0}}}.{q \over a}$$
Potential at point B,
$${V_B} = {1 \over {4\pi {\varepsilon _0}}}.{q \over b}$$
Work done in taking a charge Q from A to B,
$$W = Q({V_B} - {V_A}) = {{Qq} \over {4\pi {\varepsilon _0}}}\left[ {{1 \over b} - {1 \over a}} \right] = {{Qq} \over {4\pi {\varepsilon _0}}}\left[ {{{a - b} \over {ab}}} \right]$$

Answer

$${{qQ} \over {4\pi {\varepsilon _0}}}\left( {{{a - b} \over {ab}}} \right)$$

Answer

$${{qQ} \over {4\pi {\varepsilon _0}}}\left( {{{b - a} \over {ab}}} \right)$$

Answer

$${{qQ} \over {4\pi {\varepsilon _0}}}\left( {{b \over {{a^2}}} - {1 \over b}} \right)$$

Answer

$${{qQ} \over {4\pi {\varepsilon _0}}}\left( {{a \over {{b^2}}} - {1 \over b}} \right)$$

Electrostats & Capacitance

A charge +q is placed at the origin O of XY-axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is

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