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(II) Three point charges are arranged at the corners of a square of side l as shown in Fig. 17鈥39. What is the potential at the fourth corner (point A)?

FIGURE 17鈥39 Problem 22.

Short Answer

Expert verified

The total potential at the fourth corner or point A is \(\frac{{\sqrt 2 kQ}}{{2L}}\left( {\sqrt 2 + 1} \right)\).

Step by step solution

01

Formula for electric potential

The electric potential energy per unit charge at any point in space is termed as electric potential at that point.

The formula for the electric potential at a distance r from a point charge Q is as follows:

\(V = \frac{{kQ}}{r}\)

Here, k is the Coulomb鈥檚 constant.

02

Given information

The length of the side of a square is l.

The charges at the three corners are \( + 3Q\), \( + Q\) and \( - 2Q\).

03

Evaluation of the total potential at the fourth corner

The schematic diagram for the problem can be drawn as:

The expression for the potential at point A due to point charge\( + Q\)can be written as:

\({V_1} = \frac{{kQ}}{{\sqrt 2 l}}\)

The expression for the potential at point A due to point charge\( - 2Q\)can be written as:

\(\begin{aligned}{V_2} &= \frac{{k\left( { - 2Q} \right)}}{l}\\{V_2} &= \frac{{ - 2kQ}}{l}\end{aligned}\)

The expression for the potential at point A due to point charge\( + 3Q\)can be written as:

\(\begin{aligned}{V_3} &= \frac{{k\left( { + 3Q} \right)}}{l}\\{V_3} &= \frac{{3kQ}}{l}\end{aligned}\)

The total potential at point A can be calculated as:

\(\begin{aligned}{V_{\rm{A}}} &= {V_1} + {V_2} + {V_3}\\{V_{\rm{A}}} &= \left( {\frac{{kQ}}{{\sqrt 2 l}}} \right) - \left( {\frac{{2kQ}}{l}} \right) + \left( {\frac{{3kQ}}{l}} \right)\\{V_{\rm{A}}} &= \frac{{kQ}}{l}\left( {\frac{1}{{\sqrt 2 }} - 2 + 3} \right)\\{V_{\rm{A}}} &= \frac{{kQ}}{l}\left( {1 + \frac{1}{{\sqrt 2 }}} \right)\end{aligned}\)

Solve further as,

\(\begin{aligned}{V_{\rm{A}}} &= \frac{{kQ}}{l}\left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 }}} \right)\\{V_{\rm{A}}} &= \frac{{kQ}}{{\sqrt 2 l}}\left( {\sqrt 2 + 1} \right)\\{V_{\rm{A}}} &= \frac{{\sqrt 2 kQ}}{{2L}}\left( {\sqrt 2 + 1} \right)\end{aligned}\)

Thus, the total potential at point A is \(\frac{{\sqrt 2 kQ}}{{2L}}\left( {\sqrt 2 + 1} \right)\).

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