A uniformly charged circular arcAB of radius R is shown in the figure. It covers

Question

A uniformly charged circular arcAB of radius

R is shown in the figure. It covers a quarter

of a circle and it is located inthe second

quadrant. The total charge onthe arc is

Q >0.

The value of the Coulombconstant is

8.99 × 109 N · m2/C2.

The direction of the electric fieldvector ~E

at the origin, due to the chargedistribution,

is

1.along the negative y-axis

2.along the positive y-axis

3.in quadrant II

4.along the positive x-axis

5.in quadrant I

6.in quadrant III

7.along the negative x-axis

8. in quadrant IV

Find Ex, the x-component of the electricfield

at the origindue to the full arc length for the

case,where Q = 2.4 C and R = 1.84 m.

Answer in units ofN/C.

Explanation / Answer

First Question. the answer is in Quadrant 4 so choice 8 The electric ¯eld for a positivecharge is directed away from it. In this case, the electric fieldgenerated by each ¢qwill be directed into quadrant IV, so the totalelectric ¯eld will be in the same quadrant. I am n ot sure about the second one but Ithink you need to use the formula E field = kq/(r^2) k is the columb constant, R is radius, andq is charge. Just plug the numbers in to get the answer. I am notsure if this is correct but am sure the first question is quadrant4 I am n ot sure about the second one but Ithink you need to use the formula E field = kq/(r^2) k is the columb constant, R is radius, andq is charge. Just plug the numbers in to get the answer. I am notsure if this is correct but am sure the first question is quadrant4

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