Electric Charges And Fields

Q: A spherical conductor of radius 10 cm has a charge of 3.2 \times 10^{-7} C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? \left(\dfrac{1}{4\pi \in_0} = 9\times 10^9 Nm^/C^2\right)

The charge present inside the sphere is q=3.2\times10^{-7}C The electric field is to be determined at a point at r=15\ cm Since the point is outside the sphere, the electric field can be obtained by considering the point charge of same magnitude at the distance of 15\ cm from the sphere as the whole charge is concentrated at the center.The electric field is given by: E=\dfrac{1}{4\pi\epsilon_o}\dfrac{q}{r^{2}} Substitute the values: E=9\times 10^9\times\dfrac{3.2\times 10^{-7}}{0.0225} E=1.28\times10^5NC^{-1}

Q: Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

F=\dfrac{KQ^2}{r^2} If 25% of charge of A transferred to B then q_A = Q - \dfrac{Q}{4} = \dfrac{3 Q}{4} and q_B = -Q + \dfrac{Q}{4} = \dfrac{-3Q}{4} F_1 = \dfrac{kq_A q_B}{r^2} F_1 = \dfrac{k \left(\dfrac{3Q}{4} \right)^2}{r^2} F_1 = \dfrac{9}{16} \dfrac{kQ}{r^2} \implies F_1 = \dfrac{9F}{16}

Q: A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

Charge Q will be distributed over the surface of hollow metal sphere. (i) For r R (out side) \oint \bar{E}_0 . \overline{dS} = \dfrac{q_{en}}{\varepsilon_0} Here, q_{en} = Q (\because q_{en} = Q) \therefore E_0 4 \pi r^2 = \dfrac{Q}{\varepsilon _0} \therefore E_0 \propto \dfrac{1}{r^2}

Q: An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is?

Hint:The electron falls from rest, so, its initial velocity is zero. The electric field is present in the region, so the electron will experience electric force. \textbf{Step1: Calculation of acceleration} Let,The vertical distance to be covered is h. The electric field magnitude is E and it is directed upwards. Since the electron is negatively charged, it will experience force downwards.Electric force =q E F=e E Using newton's second law, \Rightarrow m_{e} a=e E \Rightarrow a=\frac{e E}{m_{e}} . \textbf{Step2: Calculation of time} Now using second equation of motion we get, \Rightarrow h=u t+\frac{a t^{2}}{2} \Rightarrow h=0+\frac{a t^{2}}{2} \Rightarrow h=0+\frac{a t^{2}}{2} \Rightarrow h=\frac{a t^{2}}{2} \Rightarrow h=\frac{\frac{e E}{m_{e}} \times t^{2}}{2} \therefore h=\frac{e E t^{2}}{2 m_{e}} Finding time from this, t=\sqrt{\frac{2 m_{e} h}{e E}} Now the proton takes the place of a neutron and the mass of the proton is greater than the mass of the electron but the magnitude of charge is the same. From eq (1) it is clear that t \propto \sqrt{m} , so the time taken by the proton will be greater than taken by the electron.Therefore, the correct option is 'B' is Smaller.

Q: Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+\Delta e) . If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then \Delta e is of the order of [Given mass of hydrogen m_h=1.67\times 10^{-27}Kg ]

Given : m_h= 1.67\times 10^{-27} \ kg Net charge on hydrogen atom q = (e+\Delta e)-e = \Delta e Let the distance between them be d .Equating gravitational force and the electrostatic force, we get \dfrac{Gm_h^2}{d^2} = \dfrac{k (\Delta e)^2}{d^2} Or \dfrac{(6.67\times 10^{-11})(1.67\times 10^{-27})^2}{d^2} = \dfrac{(9\times 10^9)(\Delta e)^2}{d^2} \implies \ \Delta e = 2.06\times 10^{-37} \ C The correct answer is option D.

Q: A dipole of moment p is placed in uniform electric field E then torque acting on it is given by : -

Force on charge + q = {F_A} = q E along direction of vector \vec E Force on charge - q = {F_B} = qE opposite to direction of \vec E As there two forces are equal, they from a couple \begin{array}{l} Torque\left( \tau \right) =Force\times Arm\, \, of\, \, couple \\ \tau =F\times AB\sin \theta \\ \tau =qE \times AB\sin \theta \\ \tau =qE \times 2a\sin \theta \\ \tau =\left( { q\times 2a } \right) E \sin \theta \\ \tau =PE \sin \theta ,where\, \, P\, \, in\, \, dipole\, \, movement \\ \tau =P\times E\end{array}

Q: An electric dipole is placed at an angle of 30^o with an electric field intensity 2\times 10^5\ N/C . It experiences a torque equal to 4\ Nm. The charge on the dipole, if the dipole length is 2\ cm , is :

\textbf{Step 1: Dipole moment of dipole} Given the dipole length l = 2\ cm = 2/100\ m So, the Dipole moment of dipole will be P=lq =(2/100)q P=0.02q \textbf{Step 2: Torque acting on dipole } \vec \tau = \vec P \times \vec E \Rightarrow \ \ \tau=PE\sin\theta Or, 4\ Nm=(0.02q)(2\times 10^5\ N/C)\sin30^o Or, q=200 \times 10^{-5}C =2 mC Hence, Option C is correct

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Previous Year Questions

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A spherical conductor of radius 10 cm has a charge of

3.2 \times 1 0^{- 7} C

distributed uniformly. What is the magnitude of electric field at a point

15 c m

from the centre of the sphere?

(\frac{1}{4 π \in _{0}} = 9 \times 1 0^{9} N m^{/} C^{2})

1.28 \times 1 0^{5} N / C

1.28 \times 1 0^{6} N / C

1.28 \times 1 0^{7} N / C

1.28 \times 1 0^{4} N / C

View solution

Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

F

\frac{9 F}{1 6}

\frac{1 6 F}{9}

\frac{4 F}{3}

View solution

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

Increases as r increases for

r < R

and for

r > R

Zero as r increases for

r < R

, decreases as r increases for

r > R

Zero as r increases for

r < R

, increases as r increases for

r > R

Decreases as r increases for

r < R

and for

r > R

View solution

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is?

10

times greater

Smaller

Equal

5

times greater

View solution

Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is

(e + Δ e)

. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then

Δ e

is of the order of [Given mass of hydrogen

m_{h} = 1.67 \times 1 0^{- 27} K g

]

1 0^{- 47} C

1 0^{- 20} C

1 0^{- 23} C

1 0^{- 37} C

View solution

A dipole of moment p is placed in uniform electric field E then torque acting on it is given by : -

τ = P \cdot E

τ = P \times E

τ = P + E

τ = P - E

View solution

An electric dipole is placed at an angle of

3 0^{o}

with an electric field intensity

2 \times 1 0^{5} N / C

. It experiences a torque equal to

4 N m .

The charge on the dipole, if the dipole length is

2 c m

, is :

7 μ C

8 m C

2 m C

5 μ C

View solution

Electric field at center

O

of semicircle of radius

a

having linear charge density

λ

is given as :

\frac{2 λ}{ϵ _{o} a}

\frac{λ π}{ϵ _{o} a}

\frac{λ}{2 π ϵ _{o} a}

\frac{λ}{π ϵ _{o} a}

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Two identical charged spheres suspended from a common point by two massless strings of lengths

l

, are initially at a distance

d

(d ≪ l)

apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity

v

. Then

v

varies as a function of the distance

x

between the spheres as:

v \propto x^{\frac{1}{2}}

v \propto x

v \propto x^{- \frac{1}{2}}

v \propto x^{- 1}

View solution