IP Point charges +4.7 ? C and -2.4 ? C are placed on the x axis at (17m , 0) and

Question

IP Point charges +4.7?C and -2.4?C are placed on the x axis at (17m , 0) and (-17m , 0), respectively.

Part A

There is one point on the x axis between the two charges where the potential vanishes. Is this point closer to the +4.7?C charge or closer to the -2.4?C charge?

on the halfway between them

Part B

Explain.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

Part C

Find the point referred to in part A.

on the halfway between them

Part B

Explain.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

Explanation / Answer

Part A)

Closer to the -2.4uC Charge

Part B)

Since the formula for potential is V = kq/r and potential adds algebraically, the potential from the smaller charge must cancel the potential from the larger charge. Since we are dividing by the distance, we need to divide by a larger distance from the larger charge and a smaller distance from the smaller charge in order for the two resulting numbers to completely cancel.

Part C)

kq/r = kq/r (k cancels)

(4.7)/x = (2.4)/(34 - x)

Cross multiply

159.8 - 4.7x = 2.4x

x = 22.5 m from the larger charge

That point is 22.5 - 17 = 5.51

So the coordinate is (-5.51, 0)

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