Electric Charges and Fields Test

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Electric Charges and Fields Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

An electron and a proton in a uniform electric field, the ratio of their acceleration will be

A
0

B
Unity

C
The ratio of masses of proton and electron

D
The ratio of masses of electron and proton

Solution

\(\frac{a_e}{a_m} = \frac{\frac{F_e} {m_e}}{\frac{F_p}{m_p}} \)

= \(\frac{m_p(eE)}{m_e(eE)} = \frac{m_p}{m_e}\)
Question 2
1 / -0

If E is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

A
\(E\)

B
\(E^2\)

C
\(\frac{1}{E^2}\)

D
\(E^3\)

Solution

Electrostatic energy density

\(\frac{dU}{dV} = \frac{1}{2}K \varepsilon_0E^2\)

\(\therefore \frac{dU}{dV} \propto E^2\)
Question 3
1 / -0

At a certain distance from a point charge, the electric field is 500 V/m and the potential is 3000 V. What is this distance?

A
6 m

B
12 m

C
36 m

D
144 m

Solution

Electric field due to a point charge, E = 500 V/m

Potential due to the point charge, V = 3000 V

The relation between the electric field and the electric potential is given by:

\(E = \frac{V}{d}\)

d is the distance.

d = \(\frac{V}{E}\)

d = \(\frac{3000 V}{500 V/m}\)

= 6 m

Hence the distance is 6 m
Question 4
1 / -0

A charge of 5 C experiences a force of 5000 N when it is kept in a uniform electric field. What is the P.d between two points separated by a distance of 1 cm

A
10 V

B
250 V

C
1000 V

D
2500 V

Solution

F = QE = \(\frac{QV}{d} \)

⇒ 5000 = \(\frac{5 \times V}{10^{-2}}\)

⇒ V = 10 volt
Question 5
1 / -0

The unit of intensity of electric field is

A
\(Newton/Coulomb\)

B
\(Joule/Coulomb\)

C
\(Volt-metre\)

D
\(Newton/metre\)

Solution

The SI units of the electric field are newtons per coulomb (N/C), or volts per meter (V/m).
Question 6
1 / -0

The electric field intensity inside hollow spherical Conductor is:

A
\(\frac{\sigma}{\varepsilon_0}\)

B
\(\sigma\varepsilon_0\)

C
\(Zero\)

D
\(\frac{\varepsilon_0}{\sigma}\)

Solution

A spherical Gaussian surface which lies just inside the conducting shell. Now, the Gaussian surface encloses no charge, since all of the charges lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field of the shell is zero.
Question 7
1 / -0

The dimensional formula for electric flux is

A
\([ML^2T^2]\)

B
\([ML^3T^{-3}A^{-1}]\)

C
\([ML^{-3}T^3A^1]\)

D
\([ML^{-1}T^3A^{-2}]\)

Solution

Electric flux, \(\phi = \vec E . \vec S\)

Taking dimensions both sides, we get

\([\phi] = [MLT^{-3}A^{-1}][L^2]\)

= \([ML^3T^{-3}A^{-1}]\)
Question 8
1 / -0

Gauss’s law is true only if force due to charge varies as

A
\(r^{-1}\)

B
\(r^{-2}\)

C
\(r^{-3}\)

D
\(r^{-4}\)

Solution

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula,

\(\phi = \oint \vec E . d \vec s = \frac{Q_{enc}}{\varepsilon_0}\)

No need to be a real physical surface.

\(Q_{enc}\) = charge enclosed by closed surface. It follows inverse square law.

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Re-Attempt Weekly Quiz Competition

Electric Charges and Fields Test - 19

Result

Electric Charges and Fields Test - 19

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SHARING IS CARING

Question 1

0

Unity

The ratio of masses of proton and electron

The ratio of masses of electron and proton

Solution

Question 2

\(E\)

\(E^2\)

\(\frac{1}{E^2}\)

\(E^3\)

Solution

Question 3

6 m

12 m

36 m

144 m

Solution

Question 4

10 V

250 V

1000 V

2500 V

Solution

Question 5

\(Newton/Coulomb\)

\(Joule/Coulomb\)

\(Volt-metre\)

\(Newton/metre\)

Solution

Question 6

\(\frac{\sigma}{\varepsilon_0}\)

\(\sigma\varepsilon_0\)

\(Zero\)

\(\frac{\varepsilon_0}{\sigma}\)

Solution

Question 7

\([ML^2T^2]\)

\([ML^3T^{-3}A^{-1}]\)

\([ML^{-3}T^3A^1]\)

\([ML^{-1}T^3A^{-2}]\)

Solution

Question 8

\(r^{-1}\)

\(r^{-2}\)

\(r^{-3}\)

\(r^{-4}\)

Solution

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