Q. No.The ratio of the electric flux linked with shell \(A\) and shell \(B\) in the diagram shown below is:
| 1. | \(1: 1\) | 2. | \(1: 2\) |
| 3. | \(1: 4\) | 4. | \(4: 2\) |
A square surface of a side \(L\) \(\text{(m)}\) is in the plane of the paper. A uniform electric field \(\vec{E}\) \(\text{(V/m)},\) also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is:
| 1. | \(EL^2/ ( 2ε_0 )\) | 2. | \(EL^2 / 2\) |
| 3. | zero | 4. | \(EL^2\) |
Two parallel infinite line charges with linear charge densities \(+\lambda~\text{C/m}\) and \(+\lambda~\text{C/m}\) are placed at a distance \({R}.\) The electric field mid-way between the two line charges is:
| 1. | \(\frac{\lambda}{2 \pi \varepsilon_0 {R}}~\text{N/C}\) | 2. | zero |
| 3. | \(\frac{2\lambda}{ \pi \varepsilon_0 {R}} ~\text{N/C}\) | 4. | \(\frac{\lambda}{ \pi \varepsilon_0 {R}}~\text{N/C}\) |
The electric field in a certain region is acting radially outward and is given by \(E=Aa.\) A charge contained in a sphere of radius \(a\) centered at the origin of the field will be given by:
| 1. | \(4 \pi \varepsilon_{{o}} {A}{a}^2\) | 2. | \(\varepsilon_{{o}} {A} {a}^2\) |
| 3. | \(4 \pi \varepsilon_{{o}} {A} {a}^3\) | 4. | \(\varepsilon_{{o}} {A}{a}^3\) |
According to Gauss's law in electrostatics, the electric flux through a closed surface depends on:
| 1. | the area of the surface |
| 2. | the quantity of charges enclosed by the surface |
| 3. | the shape of the surface |
| 4. | the volume enclosed by the surface |
Refer to the arrangement of charges in the figure and a Gaussian surface of a radius \(R\) with \(Q\) at the centre. Then:
| (a) | total flux through the surface of the sphere is \(\frac{-Q}{\varepsilon_0}.\) |
| (b) | field on the surface of the sphere is \(\frac{-Q}{4\pi \varepsilon_0 R^2}.\) |
| (c) | flux through the surface of the sphere due to \(5Q\) is zero. |
| (d) | field on the surface of the sphere due to \(-2Q\) is the same everywhere. |
Choose the correct statement(s):
| 1. | (a) and (d) | 2. | (a) and (c) |
| 3. | (b) and (d) | 4. | (c) and (d) |
A sphere encloses an electric dipole with charges \(\pm3\times10^{-6}~\text C.\) What is the total electric flux through the sphere?
1. \(-3\times10^{-6}~\text{N-m}^2/\text C\)
2. zero
3. \(3\times10^{-6}~\text{N-m}^2/\text C\)
4. \(6\times10^{-6}~\text{N-m}^2/\text C\)
A point charge \(q\) is placed at the center of the open face of a hemispherical surface as shown in the figure. The flux linked with the surface is:
1. zero
2. \(\frac{q}{2\varepsilon_0}\)
3. \(\frac{q}{\varepsilon_0}\)
4. \(q \pi r^2\)
| 1. | the electric field inside the surface is necessarily uniform. |
| 2. | the number of flux lines entering the surface must be equal to the number of flux lines leaving it. |
| 3. | the magnitude of electric field on the surface is constant. |
| 4. | all the charges must necessarily be inside the surface. |
| List-I (Application of Gauss Law) |
List-II (Value of \(|E|\)) |
||
| \(\mathrm{(A)}\) | The field inside a thin shell | \(\mathrm{(I)}\) | \( \dfrac{\lambda}{2 \pi \varepsilon_0 r} \hat{n} \) |
| \(\mathrm{(B)}\) | The field outside a thin shell | \(\mathrm{(II)}\) | \( \dfrac{q}{4 \pi \varepsilon_0 R^2} \hat{r} \) |
| \(\mathrm{(C)}\) | The field of thin shell at the surface | \(\mathrm{(III)}\) | \( \dfrac{q}{4 \pi \varepsilon_0 r^2} \hat{r}\) |
| \(\mathrm{(D)}\) | The field due to a long charged wire | \(\mathrm{(IV)}\) | zero |
Choose the correct answer from the options given below:
| 1. | \(\mathrm{A-IV, B-III, C-I, D-II}\) |
| 2. | \(\mathrm{A-I, B-II, C-III, D-IV}\) |
| 3. | \(\mathrm{A-IV, B-III, C-II, D-I}\) |
| 4. | \(\mathrm{A-I, B-III, C-II, D-IV}\) |
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