The figure shows some of the electric field lines corresponding to an electric field. The figure suggests
(1) EA > EB > EC
(2) EA = EB = EC
(3) EA = EC > EB
(4) EA = EC < EB
A hollow insulated conducting sphere is given a positive charge of 10μC. What will be the electric field at the centre of the sphere if its radius is 2 meters
(1) Zero
(2) 5 μCm–2
(3) 20 μCm–2
(4) 8 μCm–2
Point charges +4q, –q and +4q are kept on the x-axis at points x = 0, x = a and x = 2a respectively, then:
(1) only -q is in stable equilibrium.
(2) none of the charges are in equilibrium.
(3) all the charges are in unstable equilibrium.
(4) all the charges are in stable equilibrium.
1. | \(8~\text{L}\) | 2. | \(4~\text{L}\) |
3. | \(2~\text{L}\) | 4. | \(\frac{\text{L}}{4}\) |
Three infinitely long charge sheets are placed as shown in the figure. The electric field at point P is
(1)
(2)
(3)
(4)
Electric field at a point varies as r0 for
(1) An electric dipole
(2) A point charge
(3) A plane infinite sheet of charge
(4) A line charge of infinite length
Eight dipoles of charges of magnitude \((e)\) are placed inside a cube. The total electric flux coming out of the cube will be:
1. \(\frac{8e}{\epsilon _{0}}\)
2. \(\frac{16e}{\epsilon _{0}}\)
3. \(\frac{e}{\epsilon _{0}}\)
4. zero
A charge q is placed at the centre of the open end of the cylindrical vessel. The flux of the electric field through the surface of the vessel is
(1) Zero
(2)
(3)
(4)
\(q_1, q_2,q_3~\text{and}~q_4\) are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to the Gauss’s law?
1. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3\right) \cdot d \vec{A}=\frac{q_1+q_2+q_3}{2 \varepsilon_0}\)
2. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3\right)}{\varepsilon_0}\)
3. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}\)
4. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}\)