CALC A semicircle of radius a is in the first and second quadrants, with the center…
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Question “CALC A semicircle of radius a is in the first and second quadrants, with the center…”
CALC A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the semicircle (Fig. P21.86). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?
Answer
At the origin, electric filed:
Similar charges repel one another.
Different charges attract one another.
Each component of sin will cancel the other.
Electric filed because of small amount of charge dQ
Electric filed because of small amount of charge
From,
The strength of the electric field at origin.
The direction of the origin’s electric field.
The net field runs along the positive
d E_{\text {Total }}=2 d E \cos \theta
d E_{\text {Total }}=2 d E \cos \theta
\mathrm{dQ}
d E=\frac{k(d Q)}{a^{2}}
d Q=\lambda a d \theta
\lambda=
d E_{\text {Total }}=2 d E \cos \theta
E_{\text {Total }}=\int_{0}^{\frac{\pi}{2}} 2 \frac{k \lambda a}{a^{2}} \cos \theta d \theta
\int_{0}^{\frac{\pi}{2}} \cos \theta d \theta=[\sin \theta]_{0}^{\frac{\pi}{2}}=1
E_{\text {Total }}=\frac{2 k \lambda a}{a^{2}}
Q=\lambda a \int_{0}^{\frac{\pi}{2}} d \theta=\lambda a \frac{\pi}{2}
\lambda a=\frac{2 Q}{\pi}
E_{\text {Total }}=\frac{4 Q k}{\pi a^{2}}
x
E_{\text {Total }}=\frac{4 Q k}{\pi a^{2}} \hat{\imath}
Conclusion
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