Question

7. Four point charges are at the corners of a square of side a as shown. (a) Determine the magnitude and direction of the electric force on the charge 39. ( q-4.6 pC and a-15 cm)

Answer 1

Force on charge 3q by 2q = k(3q)(2q) /a^2

= 9*10^9*6*6*(4.6*10^-6)^2 / 0.15^2

= 50.784 N ( -ve y-direction )

Force on charge 3q by 4q = k(3q)(4q) /a^2

= 9*10^9*12*(4.6*10^-6)^2 / 0.15^2

= 101.568 N ( -ve x-direction )

Force on charge 3q by q = K*3q*q / (sqrt 2 a) ^2

= 9*10^9*3*(4.6*10^-6)^2 / 2*(0.15)^2

= 12.696 N ( angle 45^o )

Net force in X-direction = 101.568 + 12.696 cos 45 = 110.5 N

Net force in Y-direction = 50.784 + 12.696 sin 45 = 59.76 N

net force on charge 3q = sqrt(110.5^2 + 59.76^2) = 125.6 N

direction = tan^-1 ( 59.76 / 110.5) = 28.4⁰

direction = (90 + 28.4) = 118.4^o below the horizontal in clockwise direction