Shows five electric charges. Four charges with the magnitude of the charge 2.0 n

Question

Shows five electric charges. Four charges with the magnitude of the charge 2.0 nC form a square with the size a = 2.0 cm. Positive charge with the magnitude of q = 7.0 nC is placed in the center of the square. What is the magnitude of the force on the 7.0 nC charge in the middle of the figure due to the four other charges? Express your answer to two significant figures and include the appropriate units. What is the direction of the force on the 7.0 nC charge in the middle of the Figure due to the four other charges?

Explanation / Answer

The y component of the net force on 7nC due to other chares sums to zero.

Only X component of net force will have some value.

We know that, length of diagonal of the square is:

d = sqrt (a^2 + a^2) = sqrt(2)xa = 1.414 x 2 = 2.83 cm

distance of 7nC from other charges:

r = d/2 = 2.83/2 = 1.42 = 0.0142 m

Force due to the charge at left top corner is:

F = k 7nC x (-2 nC)/(0.0142)^2

F = 9 x 10^9 x 7 x 10^-9 x -2 x 10^-9/0.00021 = -6 x 10^-4 N

x component of Fx will be

Fx = -6 x 10^-4 N x cos45 = -4.24 x 10^-4 N

Since, the magnitude of all four charges is same,

Fnet-x = 4 x Fx = 4 x -4.24 x 10^-4 = -1.69 x 10^-3 N

Since Fy = 0 ; Fnet = sqrt (Fnet-x^2 + Fy^2) gives:

Hence, Fnet = 1.69 x 10^-3 N

theta = tan-1 (Fy/Fx)

but Fy = 0

hence, theta = negative X direction.

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