Question

Calculate the electric potential at the center of the square of side root 2 having charges 100uc. -50uc. 20uc and -60uc at the four corners of the square

Answer 1

potential at O due to A {V1}=K×100×10^-6/1=9×10^5
similarly V2 = 9×10^9×(-50)×10^-6/1=. -45×10^4
by this method we get V3 and V4
so, V = v1+v2+v3+v4.......
and we get the answer......
hope u get my point..........
plzz like and click on brilliant.......thank u

Answer 2

Given;-

Q1 = 100 uc

Q2 = -50 uc

Q3 = 20 uc

Q4 = -60 uc

side of square = √2 m

To Find;-

Electric potential at the center

Solution;-

We know that electric potential for a system of 4 charges is given by;-

$V = k (( \frac{q1}{r1}) + ( \frac{q2}{r2}) + ( \frac{q3}{r3}) + ( \frac{q4}{r4}))$

since, all the charges are placed on the corner of the square

So, r1 = r2 = r3 = r4 = r (let)

r = 1/2(length of diagonal of the square)

= 1/2(√2.√2) = 1 m

Hence,

$V = \frac{k}{r} (q1 + q2 + q3 + q4)$

$= \frac{k}{r} (100 - 50 + 20 - 60)$

$= \frac{9 \times {10}^{9}}{1} \times 10$

$= \frac{9 \times {10}^{10}}{ {10}^{6} } V$

$= 9 \times {10}^{4} V$

Hence, the net electric potential at the center of the square $= 9 \times {10}^{4} V$