Question

Four point charges are located at the corners of a square of side 10 cm what is force on a charge of 1uc placed a centre of square

Answer 1

Your question seems to be incorrect. Here is the correct question,

Question:Four point charges qA = 2 µC, qB = −5 µC, qC = 2 µC, and qD = −5 µC are located at the corners of a square ABCD of side 10 cm.What is the force on a charge of 1 µC placed at the centre of the square?

Answer:

First of all, we have to draw the figure(see attached document).

Then we can apply Coulomb’s law to calculate electric force on 1 µC  due to other 4 point chargers.

F = (1/4πε0)*(q1q2/r²)

Where, q1 = first charge

           q2 = Second charge

              r = distance between both charge

            ε0= permittivity of free space

But, for this question ε0 is a constant. Therefore 1/4πε0 is also a constant.

Then we can change above equation as below,

$F = K*(q1q2/r^{2} )$

$F_{A}  = K*(2/OA^{2} )$-------- (1)

$F_{B}  = K*(-5/OB^{2} )$--------(2)

$F_{C}  = K*(2/OC^{2} )$---------(3)

$F_{D}  = K*(-5/OD^{2} )$--------(4)

We know that OA= OB=OC=OD

Hence $F_{A} = F_{C} \\F_{B} = F_{D}$

But Fa, Fc & Fb,Fd are in opposite directions. (see document)

Hence sum of all forces will =0

Answer :force on a charge of 1uc = 0