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GUJCET · Physics · Electric Charges and Fields
A circle of radius '\(a\)' has charge density given by \(\lambda=\lambda_e \cos ^2 \theta\) on its circumference. What will be the total charge on the circle.
- A \(\pi a \lambda_0\)
- B zero
- C \(2 \pi a\)
- D None of these.
Answer & Solution
Correct Answer
(A) \(\pi a \lambda_0\)
Step-by-step Solution
Detailed explanation
(A)

→ As shown in figure let us take an element of angular width \(d \theta\) at angle \(\theta\). Length of this element \(=a d \theta\) charge on this element \(=d q\)
\(
d q=\lambda a d \theta=\lambda_0 \cos ^2 \theta a d \theta
\)
→Total charge \(Q =\oint d q\)
\(\begin{array}{l}\therefore Q=\int_0^{2 \pi} \lambda_0 \cos ^2 \theta a d \theta \\ \therefore Q=a \lambda_0 \int_0^{2 \pi} \cos ^2 \theta d \theta \\ \therefore Q=\pi a \lambda_0\end{array}\)

→ As shown in figure let us take an element of angular width \(d \theta\) at angle \(\theta\). Length of this element \(=a d \theta\) charge on this element \(=d q\)
\(
d q=\lambda a d \theta=\lambda_0 \cos ^2 \theta a d \theta
\)
→Total charge \(Q =\oint d q\)
\(\begin{array}{l}\therefore Q=\int_0^{2 \pi} \lambda_0 \cos ^2 \theta a d \theta \\ \therefore Q=a \lambda_0 \int_0^{2 \pi} \cos ^2 \theta d \theta \\ \therefore Q=\pi a \lambda_0\end{array}\)
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