Question

Four point charges are placed at the corners of a square of side a as shown in the figure. A charge q is slowly displaced from P₁ to P₂, then work done in

displacing the charge​

Answer 1

First, let's understand the line "q is slowly displaced from".

• Slowly displaced = change in KE is zero

Now, we know that:

$\implies \bf Work_{all} = \Delta KE$

$\rightarrow \rm W_{ext} +W_{elec} = 0$

$\rightarrow \rm W_{ext} = -W_{elec}$

•We also know that:

$\implies \bf W_{elec} = -q\Delta V$

so here work done by external will be;

$\rightarrow \rm W_{ext} = q\Delta V$

____________

☆Solution:

Refer the solution with attached image.

•First calculate the potential at P1 and P2

Potential at P1

$\longrightarrow \rm V_{at\: E} = V_A + V_B + V_C + V_D$

$\longrightarrow \rm V_{at\: E} = \dfrac{-KQ}{AE} + \dfrac{KQ}{BE} + \dfrac{-KQ}{CE}+ \dfrac{-KQ}{DE}$

$\longrightarrow \rm V_{at\: E} = \dfrac{-KQ}{\dfrac{\sqrt{5}a}{2}}+ \dfrac{KQ}{\dfrac{\sqrt{5}a}{2}}+ \dfrac{-KQ}{\dfrac{a}{2}}+ \dfrac{KQ}{\dfrac{a}{2}}$

$\longrightarrow \rm V_{at\: E} = 0 + 0 =0$

》$\bf Potential\: at\: P_1 = 0$

Similarly by above method;

》$\bf Potential\: at\: P_2 = 0$

---------

$\rightarrow \rm W_{ext} = q(0-0) =0$

☆☆Work done to displace the charge is zero.

Answer 2

Answer:

by work done formula

Wext =q( Vp1 - Vp2 )

Vp1 = Vp2

W = 0