A charged particle is located 5 m above an infinite charged plane (indicated by
ID: 1600855 • Letter: A
Question
A charged particle is located 5 m above an infinite charged plane (indicated by the horizontal line in the figure at the top of the next page below) with a surface charge density of 25 mC per square meter. The electric potential a distance z above the plane is V(Z) = -1412 z j/(C m). Using the relation between the electric potential and the electric field, what is the electric force on the point charge if it carries charge Q = - .5 C? Suppose we now introduce a second particle with charge q = 2 nC as shown in the figure below. Using the principle of superposition, what is the total electric force on charge Q due to the 2 nC charge and the charged infinite plane? Use the x-z coordinate system in the figure below when expressing your answer for the force. Where should we place a third particle with charge equal to the second. 2 nC, such that the total force on charge Q is equal to your answer from part (a) (assume the third particle lies in the plane of the page, like the other two particles)?Explanation / Answer
13 a)
V(z) = - 1412 z
electric field is given as
E(z) = - dV(z)/dz = - (d/dz) (-1412 z) = 1412 towards positive z-direction
Q = charge = - 0.5 C
electric force is given as
Fz = Q E(z) = 0.5 x 1412 = 706 N towards negative z-direction since the charge is negative and hence experience electric force in opposite direction of electric field.
b)
r = distance between the charges = 12 m
electric force by charge "q" on charge Q is given using coulomb's law as
F = k Qq/r2 = (9 x 109) (0.5) (2 x 10-9)/(12)2 = 0.0625 N towards positive X-axis
Net force is given as
Fnet = sqrt(Fz2 + F2) = sqrt((706)2 + (0.0625)2) = 706 N
c)
to cancel out the effect of force "F" by charge "q" in positive x-direction , we must produce same force "F" in opposite negative X-direction. this can be done by placing a charge "q" at distance 12 m to the left of charge .
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