The point charges are q 1 = -3.8 C andq 2 = +2.2 C The charges are placed at (9m

Question

The point charges are q1= -3.8 C andq2= +2.2 C The charges are placed at (9m,0) and (-9m,0)respectively. Let p be a point to the left of the positive charge where theelectric potential vanishes.This is shown in the followingfigure: Find the point to the left of the positive charge where theelectric potential vanishes. note: The answer given is -33.8m but i have no idea how thatwas gotten anybody know how to do this problem?? The point charges are q1= -3.8 C andq2= +2.2 C The charges are placed at (9m,0) and (-9m,0)respectively. Let p be a point to the left of the positive charge where theelectric potential vanishes.This is shown in the followingfigure:

Explanation / Answer

The pointcharges are q1= -3.8 C and q2= +2.2C The chargesare placed at (9m,0) and (-9m,0)respectively. Let p be apoint to the left of the positive charge where the electricpotential vanishes.
Find thepoint to the left of the positive charge where the electricpotential vanishes.

{Total ElectricPotential @ P} = V =
     = {ElectricPotential From (q1) @ P} + {Electric Potential From (q2) @P}
     = (1/(4*Pi*))*(q1)/(R1)+ (1/(4*Pi*))*(q2)/(R2)
     = (8.988e+9)*(-3.8e-6)/(9.0 -x) + (8.988e+9)*(+2.2e-6)/(-9.0- x)

{V = 0}:
---->    (8.988e+9)*(-3.8e-6)/(9.0 - x) +(8.988e+9)*(+2.2e-6)/(-9.0 -x) = 0
---->    (-3.8)/(9.0 - x) +(+2.2)/(-9.0- x) = 0
---->    (-3.8)*(-9.0 - x) + (+2.2)*(9.0 - x) = 0
---->    1.6*x + 54.0 = 0
---->    x = (-54.0)/(1.6)
---->    x = -33.75 meters

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