Electric Charges and Fields - Result Question 39

39. A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of voltmeter associated with the curved surface $B$, the flux linked with the plane surface $A$ in units of voltmeter will be

[2007]

$ \text{ B } $

(a) $\frac{q}{2 \varepsilon_0}$

(b) $\frac{\phi}{3}$

(c) $\frac{q}{\varepsilon_0}-\phi$

(d) $\frac{1}{2}(\frac{q}{\varepsilon_0}-\phi)$

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Answer:

Correct Answer: 39. (d)

Solution:

  1. (d) Since $\phi _{\text{total }}=\phi_A+\phi_B+\phi_C=\frac{q}{\varepsilon_0}$,

where $q$ is the total charge.

As shown in the figure, flux associated with the curved surface B is $\phi=\phi_B$

Let us assume flux linked with the plane surfaces $A$ and $C$ be

$\phi_A=\phi_C=\phi^{\prime}$

Therefore,

$\frac{q}{\varepsilon_0}=2 \phi^{\prime}+\phi_B=2 \phi^{\prime}+\phi$ $\Rightarrow \phi^{\prime}=\frac{1}{2}(\frac{q}{\varepsilon_0}-\phi)$



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