Six point charges, four positive (+ q ) and two negative(- q ), are fixed at the

Question

Six point charges, four positive (+q) and two negative(-q), are fixed at thepositions shown in thex-y plane; q = 2.5 ×10-6 Cand a = 0.91 m. A test charge Q= 0.6 × 10-6 C is at the origin.

(a) Find the x- and y-components of the totalforce on Q:

(b) Find the x- and y-components ofthe totalelectric field at the origin due to the six q charges:

Six point charges, four positive (+q) and two negative(-q), are fixed at thepositions shown in thex-y plane; q = 2.5 ×10-6 Cand a = 0.91 m. A test charge Q= 0.6 × 10-6 C is at the origin.

(a) Find the x- and y-components of the totalforce on Q:

(b) Find the x- and y-components ofthe totalelectric field at the origin due to the six q charges:

Explanation / Answer

forces on Q due to charges at (0, a) and (0, -a) have samemagnitude and opposite directions, so cancel each other. forces on Q due to charges at (a, 0) and (-a, 0) have samemagnitude kqQ/a2 and same direction (-x), so their sumis 2kqQ/a2 (-i) forces on Q due to charges at (a, a) and (-a, -a) have samemagnitude kqQ/(2a2) and same direction (i + j), so theirsum is 2kqQ/(2a2) * (i/2 + j/2), Fx = -2kQ/a2 + kqQ/(2a2) =kqQ/a2 *(-2 + 1/2) = -0.0211 N Fy = kqQ/(2a2) = 0.0115 N electric field E = F/Q Ex = -2k/a2 + kq/(2a2) =kq/a2 *(-2 + 1/2) = -3.52*104 N/C Ey = kq/(2a2) = 1.92*104 N/C

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