Two negative charges, Q = -50 C are placed on the x-axis at x = +0.35 m and x =

Question

Two negative charges, Q = -50 C are placed on the x-axis at x = +0.35 m and x = -0.35 m.

Part A -How much energy does it take to assemble these two charges?

Part B You place an unknown charge on the y-axis at y = 0.7 m and find that the electric potential is 0 V at the origin. What is the unknown charge?

Part C If you place a -2 C charge at the origin, what is the x-component of the force on it from the other three charges?

Part D If you place a -2 C charge at the origin, what is the y-component of the force on it from the other three charges?

Explanation / Answer

let q1 = q2 = -50 micro C

distance between the two charges, d = 0.7 m

A) Energy required to assemble the charges = Potentail energy of the two charges

= k*q1*q2/d

= 9*10^9*(-50*10^-6)*(-50*10^-6)/0.7

= 32.14 J

B) let Q is the unknown charge

Vnet = 0

V1 + V2 + V = 0

V = -(V1+V2)

k*Q/y = -k*(q1+q2)/x

Q/y = -(q1+q2)/x

Q/0.7 = -(-50-50)*10^-6/0.35

Q = 200*10^-6 C

C) x-component of force will be zero.

Fx = 0

d) The net force due to q1 and q2 will be zero. only Q exerted force new charge.

Fy = k*Q*q/y^2

= 9*10^9*200*10^-6*2*10^-6/0.7^2

= 7.35 N

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