Question

Q7) Four charges are arranged at the corners of a square ABCD, as shown. The force on +ve charge kept

at the center pf the square is

a. Zero

b. Along diagonal AC

c. Along diagonal BD

d. Perpendicular to the side AB​

Answer 1

Answer:This may help you

Explanation:

The force on +Q at the center of square is,

F=k[∣ r 0− r 1∣ 3−qQ( r-r1)+ ∣ r 0−r ∣ 3

qQ( r 0− r 2​)+ ∣ r 0− r 3∣ 3−2qQ( r − r 3)

​+ ∣ r 0− r 4 ∣ 3

2qQ( r 0 − r 4)]= r

2kqQ[( i^+ j^)−(− i^+ j^)+2(− i^ − j^)−2(i^−j^)]= r

2kqQ[i^+ j^+ i^−j^−2i^−2j^−2i^+2j^]=r

2kqQ(−2i^)

thus, F is perpendicular to side AB.