Question

Four electric charges \(+q, +q, -q\) and \(-q\) are placed at the corners of a square of side \(2L\) (see figure). The electric potential at point \(A\), midway between the two charges \(+q\) and \(+q\), is

• A. $ \frac{1}{ 4 \pi \varepsilon_0 } \frac{ 2q}{ L} (1 + \sqrt 5)$
• B. $\frac{1}{ 4 \pi \varepsilon_0 } \frac{ 2q}{ L} \bigg [ 1 - \frac{ 1}{ \sqrt 5} \bigg ] $
• C. $ \frac{1}{ 4 \pi \varepsilon_0 } \bigg(1 + \frac{ 2q}{ L} (1 - \frac{1}{\sqrt 5})\bigg)$
• D. zero

Answer 1

The Correct Option is C

$V _{ A }=\frac{ kq }{ L }+\frac{ kq }{ L }-\frac{ kq }{\sqrt{5}\, L }-\frac{ kq }{\sqrt{5}\, L }$
$=\frac{2 \,kq }{ L }\left(1-\frac{1}{\sqrt{5}}\right)$

Answer 2

Here, the electric potential at any point is defined as the electric potential energy per unit charge. 

So, mathematically, we have electric potential as: \(V=\frac{U}{q}\), where q is the test charge. 

Let’s consider a point charge having charge q which is at origin. 

Now, consider another point charge q₀ , Here is at infinity. 

This charge q₀ is under the influence of the electric field of charge q. 

Now, bring the charge q₀ from infinity to position vector r. 

Eventually, work needs to be done against the electric field in order to change the configuration of the system or the position of the charge q₀ .

The force on the charge q₀ is \(F=\frac{kqq_{0}}{r^{2}}\), where \(k=\frac{1}{4\pi \varepsilon _{0}}\). 

Hence the Work done will be given as \(W=\int dW = \int_{\infty }^{r}Fdr\)

 ⇒ \(W= \int_{\infty }^{r}\frac{kqq_{0}}{r^{2}}dr\) 

⇒ \(W= kqq_{0}\int_{\infty }^{r}\frac{1}{r^{2}}dr\) 

⇒ W=kqq0 [−1/r]r∞ 

⇒ \(W= kqq_{0}[-\frac{1}{r}]_{\infty }^{r}\)

⇒ \(W= -kqq_{0}(\frac{1}{r}-\frac{1}{\infty})\)

\(W= \frac{-kqq_{0}}{r}\)

Hence, the change in potential energy is \(U_{f}-U_{i}=-W=-(-\frac{kqq_{0}}{r})\)

⇒ \(U_{f}-U_{i}=\frac{kqq_{0}}{r}\) Lets assume, the potential energy at infinity to be zero, \(\rightarrow U_{i}=0\) and

Hence, the potential energy at the position vector r is given by  \(U=\frac{kqq_{0}}{r}\)

As mentioned before, the electric potential is defined as potential energy per unit charge. 

So, potential due to a point charge at any point is given as: \(V=\frac{kq}{r}\). 

Therefore, the electric potential due to top left charge is: \(V_{1}=\frac{kq}{\frac{2L}{2}}\).

Likewise, the electric potential due to bottom left charge is: \(V_{2}=\frac{kq}{L}\). 

Therefore, the electric potential due to top right point charge will be \(V_{3}=\frac{-k(-q)}{\sqrt{(\frac{2L}{2})^{2}}+(2L)^{2}}\)=\(\frac{-kq}{\sqrt{5}L}\). 

As the bottom right point charge is exact to the top right point charge, 

Therefore, the potential due to that charge will also: \(V_{4}=-\frac{1}{\sqrt{5}}\frac{kq}{L}\). 

\(V=\sum_{i=1}^{4}V_{i}=V_{1}+V_{2}+V_{3}+V_{4}\)

⇒ \(V= \frac{kq}{L}+\frac{kq}{L}+(-\frac{1}{\sqrt{5}}\frac{kq}{L})+(-\frac{1}{\sqrt{5}}\frac{kq}{L})\)

⇒ \(V=\frac{2kq}{L}(1-\frac{1}{\sqrt{5}})\) 

As, \(k=\frac{1}{4\pi \varepsilon _{0}}\), then we have, \(V=\frac{1}{4\pi \varepsilon _{0}}\frac{2q}{L}(1-\frac{1}{\sqrt{5}})\). Therefore, then the electric potential at point A, midway between the two charges +q and +q is: \(\frac{1}{4\pi \varepsilon _{0}}\frac{2q}{L}(1-\frac{1}{\sqrt{5}})\) 

Therefore , the option C is the correct option. .