Question

A particle with charge\( + {q_1}\)and a particle with charge\( - {q_2}\)are located as shown in figure 16.91. What is the potential (relative to infinity) at location A.

Short answer

The electric potential at location A due to both charge is \(\frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}}}{{{r_{1A}}}} - \frac{{{q_2}}}{{{r_{2A}}}}} \right)\).

Step-by-step solution

Identification of given data

The positive charge on the particle is\({q_1}\).

The distance of location A from positive charge is\({r_{1A}}\).

The positive charge on the particle is\( - {q_2}\)

The distance of location A from negative charge is \({r_{2A}}\).

Conceptual Explanation

The electric potential is the effect of charged particle at some distance from the charged particle.It is a scalar quantity so the net potential due to different charges at a location is calculated by the algebraic sum of potential of each charge.

Determination of electric potential at location A

The electric potential at location A due to positive charge is given as:

\({V_1} = \frac{{{q_1}}}{{4\pi {\varepsilon _0}{r_{1A}}}}\)

The electric potential at location A due to negative charge is given as:

\({V_2} = \frac{{ - {q_2}}}{{4\pi {\varepsilon _0}{r_{2A}}}}\)

The electric potential at location A due to both charges is given as:

\(\begin{array}{c}V = {V_1} + {V_2}\\V = \frac{{{q_1}}}{{4\pi {\varepsilon _0}{r_{1A}}}} + \left( {\frac{{ - {q_2}}}{{4\pi {\varepsilon _0}{r_{2A}}}}} \right)\\V = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}}}{{{r_{1A}}}} - \frac{{{q_2}}}{{{r_{2A}}}}} \right)\end{array}\)

Therefore, the electric potential at location A due to both charge is \(\frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}}}{{{r_{1A}}}} - \frac{{{q_2}}}{{{r_{2A}}}}} \right)\).