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 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.
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
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 a", so Electric field strength at the center due one of the charges is:
E = k*magnitude of any of the charges/[Sqrt 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
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) * (k Q/(2 a^2) Cos[Pi/4]
= Sqrt k Q/a^2
these are all possibilities in fact. Now you choose your case by watching your diagram.