Question

Four +9 µC point charges are at the corners of a square of side 4 m. Find the potential at the center of the square (relative to zero potential at infinity) for each of the following conditions.(a) All the charges are positive

(b) Three of the charges are positive and one is negative
(c) Two are positive and two are negative

Answer 1

(a) 81000 V, (b) 40500 V, (c) 0 V

Step-by-step explanation:

Given:

side of square, a = 4 m

charge on each corner , q = + 9 micro coulomb

The formula for the potential is

`V = (K q)/(r)`

where, k is the Coulombic constant.

(a)

All the charges are positive:

Let V1, V2, V3 and V4 be the potential at the centre O due to the charges at 1, 2,3 and 4.

`V_(1) = (K q)/(a)`

`V_(2) = (K q)/(a)`

`V_(3) = (K q)/(a)`

`V_(4) = (K q)/(a)`

Total potential at the centre of square is

`V = V_(1)+ V_(2)+ V_(3)+ V_(4)`

`V = (4K q)/(a)`

`V = (9* 10^(9)* 4* 9* 10^(-6))/(4)`

V = 81000 V

(b)

Three charge are positive and one is negative

`V_(1) = (K q)/(a)`

`V_(2) = (K q)/(a)`

`V_(3) = (K q)/(a)`

`V_(4) = -(K q)/(a)`

Total potential at the centre of the square is

`V = V_(1)+ V_(2)+ V_(3)+ V_(4)`

`V = (2K q)/(a)`

`V = (9* 10^(9)* 2* 9* 10^(-6))/(4)`

V = 40500 V

(c)

two charge are positive and the other two are negative

`V_(1) = (K q)/(a)`

`V_(2) = (K q)/(a)`

`V_(3) = -(K q)/(a)`

`V_(4) = -(K q)/(a)`

Total potential at the centre of the square is

`V = V_(1)+ V_(2)+ V_(3)+ V_(4)`

V = 0 V

Answer 2

Answer with Explanation:

We are given that

`q_1=q_2=q_3=q_4=q=9\mu C=9* 10^(-6) C`

`1\mu C=10^(-6) C`

Side of square,d=4 m

Distance of charge from the center of square,d=
`(1)/(2)√(4^2+4^2)=2\sqrt 2m`

Potential is a scalar quantity therefore, we consider sign .

a.When all charges are positive

Then, Potential at center of square,V=
`4* (kq)/(d)`

Where
`k=9* 10^9`

`V=4* (9* 10^9* 9* 10^(-6))/(2\sqrt 2)=1.15* 10^5 V`

b.Potential at center of square,V=
`3(kq)/(d)-(kq)/(d)=(2kq)/(d)`

`V=(2* 9* 10^9* 9* 10^(-6))/(2\sqrt 2)=5.7* 10^(4) V`

c.Potential at center of square,V=
`(2kq)/(d)-(2kq)/(d)=0`