Electrostatics

A charge Q is uniformly distributed over a long rod AB of length L as shown in the Figure ….

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Q: A charge Q is uniformly distributed over a long rod AB of length L as shown in the Figure . The electric potential at the point O lying at distance L from the end A is :

$\displaystyle (a) \frac{Q}{8\pi \epsilon_0 L} $

$\displaystyle (b) \frac{3Q}{8\pi \epsilon_0 L} $

$\displaystyle (c) \frac{3Q}{4 \pi \epsilon_0 L (ln2)} $

$\displaystyle (d) \frac{Q (ln2)}{4 \pi \epsilon_0 L} $

Click to See Answer :
Ans: (d)

Sol: Consider a small element of length dx of the rod at a distance x from O .

Numerical

Charge on the element $\displaystyle dq = \frac{Q}{L}dx $

Potential at O due to this element is

$\displaystyle dV = \frac{1}{4\pi \epsilon_0} \frac{dq}{x} $

Potential at O due to entire rod is

$\displaystyle V = \int_{L}^{2L} \frac{1}{4\pi \epsilon_0} \frac{dq}{x} $

$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} \int_{L}^{2L} \frac{dx}{x} $

$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} [lnx]_{L}^{2L} $

$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} [ln 2L -lnL] $

$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} ln(\frac{2L}{L}) $

$\displaystyle V = \frac{Q (ln2)}{4 \pi \epsilon_0 L} $

 

What is the flux through a cube of side ‘ a ‘ if a point charge of q is at one of its corner :

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Q: What is the flux through a cube of side ‘ a ‘ if a point charge of q is at one of its corner :

$\displaystyle (a) \frac{2q}{\epsilon_0} $

$\displaystyle (b) \frac{q}{8 \epsilon_0} $

$\displaystyle (c) \frac{q}{\epsilon_0} $

$\displaystyle (d) \frac{q}{2\epsilon_0}6a^2 $

Click to See Answer :
Ans: (b)
Sol: We have shown a point charge q at one corner of a cube of side a. This charge can be imagined at the centre of eight identical cubes as shown .
Numerical
According to Gauss’s theorem in electrostatics electric flux through the given cube,
$\displaystyle \phi = \frac{1}{8}(\frac{q}{\epsilon_0})= \frac{q}{8\epsilon_0} $

 

A cubical region of side a has its centre at the origin. It encloses three fixed point charges ….

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Q: A cubical region of side a has its centre at the origin. It encloses three fixed point charges -q at (0,-a/4,0), + 3 q at (0,0,0) and -q at (0,a/4,0), Fig Choose the wrong option(s).

Numerical

(a) The net electric flux crossing the plane x = + a/2 is equal to the net electric flux crossing the plane x = -a/2

(b) The net electric flux crossing the plane y = + a/2 is more than the net electric flux crossing the plane y = -a/2

(c) The net electric flux crossing the entire region is q/ε_0

(d) The net electric flux crossing the plane z = a/2 is equal to the net electric flux crossing the plane x= + a/2

Click to See Answer :
Ans: (b)
Sol: Cubical region of side a having centre at(0,0,0).
It encloses three fixed point charges :
-q at A(0,-a/4),+3q at O(0,0,0) and -q at B(0,a/4,0). As+3 q is at the origin O, and charges -q,-q are placed symmetrically about planes x=+a/2 and x=-a/2, therefore, electric flux through them is same.
Choice (a) is correct.According to Gauss’s theorem , the net electric flux crossing the given region  is
φ=∑qin/∈0 =(-q+3q-q)/∈0 = q/∈0
Choice (c) is correct .

Again, as + 3 q is at the origin O, and charges -q,-q are placed symmetrically about the planes z=+a/2 and x=-a/2 so net electric flux crossing these planes is equal.
Choice (d) is correct.

The only wrong option is (b).

 

A hollow cylinder has a charge q coulomb within it. If φ is the electric flux in unit of volt-m ….

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Q: A hollow cylinder has a charge q coulomb within it. If φ is the electric flux in unit of volt-m associated with the curves surface B, the flux linked with the plane surface A in the same units will be

Numerical

(a) φ/3

$\displaystyle (b) \frac{q}{2 \epsilon_0} $

$\displaystyle (c) \frac{1}{2}(\frac{q}{\epsilon_0} -\phi) $

$\displaystyle (d)(\frac{q}{\epsilon_0} -\phi) $

Click to See Answer :
Ans: (c)

Sol: According to Gauss’s Law

Total Electric Flux ,

$\displaystyle \phi_{Total} = \frac{q}{\epsilon_0} $

$\displaystyle \phi_{Total} = \phi_A + \phi_B + \phi_C $

$\displaystyle \frac{q}{\epsilon_0} = \phi_A + \phi + \phi_C $

$\displaystyle \frac{q}{\epsilon_0} = 2 \phi_A + \phi $ ; (Since φA = φC)

$\displaystyle \frac{q}{\epsilon_0} – \phi = 2 \phi_A $

$\displaystyle \phi_A = \frac{1}{2}(\frac{q}{\epsilon_0} -\phi) $

 

A charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre….

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Q: A charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre. The potential at a point distant 15 cm from the centre will be

(a) 3 V

(b) 3/2 V

(c) 2/3 V

(d) 1/3 V

Click to See Answer :
Ans: (c)
Sol:
At any point inside a charged conductor, potential is same as on its surface, i.e,$\displaystyle V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$ , where R = 10 cm.

At r = 15 cm,

$\displaystyle V’ = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$

$\displaystyle V’ = \frac{q}{4\pi\epsilon_0 R} \times \frac{R}{r}$

$\displaystyle V’ = V \times \frac{10}{15}$

V’ = 2V/3