Four point charges ( + 4Q), (- 2Q) and (+Q) and (-Q) with Q > 0 arc placed as sh

Question

Four point charges ( + 4Q), (- 2Q) and (+Q) and (-Q) with Q > 0 arc placed as shown in the figure. Find the magnitude and direction of the net electric field at the origin O of the system of coordinates What would be the magnitude an direction of the electric force acting, on a charge (- Q ) which is placed at the origin? Repeat for a charge (+Q). If the charge (-2Q) is removed from the x axis what will be the magnitude and the direction of the electric field produced b y the remaining 3 charges at point O?

Explanation / Answer

a. Ey = -kQ/a^2 + kQ/(2a)^2 = -(3/4)*(kQ/a^2)(assuming up is positive y) Ex = k(2Q)/(2a)^2 + k(4Q)/(2a)^2 = (3/2)(kQ/a^2)( assuming right is positive x) Enet direction = < -(3/4)*(kQ/a^2) , (3/2)(kQ/a^2) > Enet magnitude = sqrt([-(3/4)*(kQ/a^2)]^2 , [(3/2)(kQ/a^2)]^2) b. For -Q: Fnet direction = Enet direction = < -(3/4)*(kQ/a^2) , (3/2)(kQ/a^2) > Fnet magnitude = -Q * Enet magnitude = -Q * sqrt([-(3/4)*(kQ/a^2)]^2 , [(3/2)(kQ/a^2)]^2) For +Q: Fnet direction = Enet direction = < -(3/4)*(kQ/a^2) , (3/2)(kQ/a^2) > Fnet magnitude = Q * Enet magnitude = Q * sqrt([-(3/4)*(kQ/a^2)]^2 , [(3/2)(kQ/a^2)]^2) c. Ey = -kQ/a^2 + kQ/(2a)^2 = -(3/4)*(kQ/a^2) (assuming up is positive y) Ex = k(4Q)/(2a)^2 = kQ/a^2 ( assuming right is positive x) Enet direction = < -(3/4)*(kQ/a^2) , kQ/a^2 > Enet magnitude = sqrt([-(3/4)*(kQ/a^2)]^2 , [kQ/a^2]^2)

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