Answer:
A uniformly charge d insulating rod of length 14.0cm is bent into the shape of a semicircle as shown in Figure. The rod has a total charge of −7.50μC. Find (a) the magnitude and (b) the direction of the electric field at O, the center of the semicircle.
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Medium
Solution
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Verified by Toppr
Due to symmetry, E
y
=∫dE
y
=0, and E
x
=−∫dEsinθ=−k
e
∫
r
2
dqsinθ
where dq=λds=λrdθ; the component E
x
is negative because charge q=−750μC, causing the net electric field to be directed to the left.
E
x
=−
r
k
e
λ
0
∫
π
sinθdθ=−
r
k
e
λ
(−cosθ)∣
0
π
=−
r
2k
e
λ
where λ=
and r=
π
L
. Thus,
E
x
=−
L
2
2k
e
∣q∣π
=−
(0.140m)
2
2(8.99×10
9
N⋅m
2
/C
2
)(7.50×10
−6
C)π
E
x
=−2.16×10
7
N/C
(a) magnitude E=
2.16×10
7
N/C