1. - The field just outside a 4.35cm -radius metal ball is 5.79×10^2N/C and poin

Question

1. - The field just outside a 4.35cm -radius metal ball is 5.79×10^2N/C and points toward the ball. What charge resides on the ball?

2. - A positive point charge Q1 = 2.9×10^-5 C is fixed at the origin of coordinates, and a negative charge Q2= -6.7×10^-6 C is fixed to the axis at x=+3.0m. Find the location of the place(s) along the axis where the electric field due to these two charges is zero.
Express your answer(s) using two significant figures. If there is more than one answer, enter each answer separated by a comma.

Explanation / Answer

E = kq/r^2
(8.99E9)(q)/(0.0435)^2 = 5.79E2
q= 1.22 E -10 C
since the field points inward, teh charge is negative

so q = -1.22 E-10 C

2)

The electric fields must equal each other in magnitude. This does not occur between the charges as both fields are in the direction. It cannot be at x< 0 as the field created by the closer, positive charge will always be stronger than the one created by the negative as the positive charge is greater in magnitude.

So thus, the only area where the field can be 0 is for x>=3.0 m as the negative charge is closer but the positive charge is bigger

kq/r^2 = kq/r^2

q/r^2 = q/r^2

2.9E-5/x^2 = 6.7E-6/(x-3)^2

Using a solver, it is found that x=2.03 m, 5.78 m

The first solution cannot be correct in a physics sense as the fields do not cancel each other out

so the electric field is 0 at x=5.78 m

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