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Engineering
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The electrostatic potential inside a charged spherical ball is given by $\dpi{100} \phi = ar^{2}+b$ where $\dpi{100} r$ is the distance from the centre;     $\dpi{100} a,b$ are constants. Then the charge density inside the ball is

• Option 1)

$-24\pi a\varepsilon _{0}r$

• Option 2)

$-6 a\varepsilon _{0}r$

• Option 3)

$-24\pi a\varepsilon _{0}$

• Option 4)

$-6 a\varepsilon _{0}$

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

$\dpi{100} \phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}$

- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.

$\phi=ar^{2}+b$

Electric field   $E=\frac{-d\phi}{dr}=-2ar$ - (i)

$\phi=\int \vec{E}.d\vec{S}=\frac{q}{\varepsilon _{0}}$

$-2ar\:4\pi r^{2}=\frac{q_{in}}{\epsilon _{0}}$

$q_{inside}=-8\varepsilon _{0}a\pi\:r^{3}$

change density inside the ball is

$\rho _{inside}=\frac{q_{inside}}{\tfrac{4}{3}\pi\:r^{3}}$

$=\frac{-8\varepsilon _{0}a\pi\:r^{3}}{\tfrac{4}{3}\pi\:r^{3}}$

$\rho _{inside}=-6a\varepsilon _{0}$

Option 1)

$-24\pi a\varepsilon _{0}r$

This is incorrect option

Option 2)

$-6 a\varepsilon _{0}r$

This is incorrect option

Option 3)

$-24\pi a\varepsilon _{0}$

This is incorrect option

Option 4)

$-6 a\varepsilon _{0}$

This is correct option

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Engineering
133 Views   |

Two identical charged spheres suspended from a common point by two massless strings of length $\dpi{100} l$ are initially a distance $\dpi{100} d(d< < l)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity $\dpi{100} \upsilon$ . Then as a function of distance $\dpi{100} x$ between them

• Option 1)

$\upsilon\: \alpha\: x^{-1/2}$

• Option 2)

$\upsilon\: \alpha\: x^{-1}$

• Option 3)

$\upsilon\: \alpha\: x^{1/2}$

• Option 4)

$\upsilon\: \alpha\: x$

Option 1 ans

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