# Four point charges of equal magnitude, are held at the corners of a square as shown below.The length of each side of the square is 2a and the sign of the charges is as shown. The point P is at the...

Four point charges of equal magnitude, are held at the corners of a square as shown below.

The length of each side of the square is 2*a* and the sign of the charges is as shown. The point P is at the centre of the square.

a) Deduce that the magnitude of the electric field strength at point P due to one of the point charges is equal to kQ/[2(a^2)]

b)Determine, in terms of *Q*, *a* and *k*, the magnitude of the electric field strength at point P.

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Let me assume the four charges to be q1, q2, q3, q4 , now in the final answer you put the signs of the charges and put Q for each. Let the square be ABCD, q1 is at A, q2 is at B, q3 is at C, q4 is at D,

remember all have same magnitude, that is

|q1|=|q2|=|q3|=|q4|=Q.

a) we need the magnitude of Electric field strength at P due to any of the charges, Now as length of each side is 2a, so distance between any vertex and the center P is "Sqrt[2] a", so Electric field strength at the center due one of the charges is:

E = k*magnitude of any of the charges/[Sqrt[2] a ]^2

= k Q/(2 a^2)

b) Now we need to vector sum the electric field strengths due to the 4 charges at P:

E = k q1/(2 a^2) along A to P or P to A

+ k q2/(2 a^2) along B to P or P to B

+ k q3/(2 a^2) along C to P or P to C

+ k q4/(2 a^2) along D to P or P to D

as you have not given the diagram, so let me write down all the possibilities:

possibility 1) if all the charges are positive,

E = 0, as "along A to P" will cancel "along C to P"

and "along B to P" will cancel "along D to P"

possibility 2) If q1 at A & q3 at C are of same type +ve/-ve, and

q2 at B & q4 at C are of same type +ve/-ve, then

again

E =0 by the same logic as that of possibility (1).

Possibility 3) If q1 & q2 are +ve and q3 & q4 are -ve,

E = k Q/(2 a^2) along A to P

+

k Q/(2 a^2) along B to P

+

k Q/(2 a^2) along P to C

+

k Q/(2 a^2) along P to D

= 2 * (k Q/(2 a^2) Cos[Pi/4] to P to center of DC

+

2 * (k Q/(2 a^2) Cos[Pi/4] to P to center of DC

magnitude of E = 4 * (1/Sqrt[2]) * (k Q/(2 a^2) Cos[Pi/4]

= Sqrt[2] k Q/a^2

these are all possibilities in fact. Now you choose your case by watching your diagram.