Question 1:
easyA spherical conductor of radius $10 \text{ cm}$ has a charge of $3.2 \times 10^{-7} \text{ C}$ distributed uniformly. What is the magnitude of electric field at a point $15 \text{ cm}$ from the centre of the sphere? (2020) $(\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2)$
Electric field $E = \frac{kq}{r^2}$. Substituting the given values, $E = \frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{(0.15)^2}$. Solving this yields $E = 1.28 \times 10^5 \text{ N/C}$.
Question 2:
easyA hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre (2019)
Inside a hollow metal sphere, the electric field is zero ($E = 0$ for $r < R$). Outside the sphere, it behaves as a point charge concentrated at the center, so $E \propto \frac{1}{r^2}$, which decreases as r increases for $r > R$.
Question 3:
easyTwo 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 : (2019)
Initial force $F = \frac{kQ^2}{r^2}$. When 25% ($Q/4$) of A is transferred to B, new charge on A is $Q - Q/4 = 3Q/4$ and on B is $-Q + Q/4 = -3Q/4$. The new force is $F' = \frac{k(3Q/4)(3Q/4)}{r^2} = \frac{9}{16}\frac{kQ^2}{r^2} = \frac{9F}{16}$.
Question 4:
easyAn 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 (2018)
The time of fall is given by $t = \sqrt{\frac{2h}{a}} = \sqrt{\frac{2hm}{qE}}$. Since the mass of an electron $m_e$ is much smaller than the mass of a proton $m_p$, the time of fall for the electron will be smaller.
Question 5:
easyThe electric field at a distance $\frac{3R}{2}$ from the center of a charged conducting spherical shell of radius R is E. The electric field at a distance $\frac{R}{2}$ from the center of the sphere is (2010 Main)
The electric field inside a charged conducting spherical shell is zero at all points. Since the distance $\frac{R}{2}$ is less than the radius R, the point lies inside the shell. Therefore, the electric field is zero.
Question 6:
easyIn Millikan oil drop experiment, a charged drop falls with a terminal velocity V. If an electric field E is applied vertically upwards it moves with terminal velocity 2V in upward direction. If electric field reduces to E/2 then its terminal velocity will be: (1990)
Downward falling: $mg = 6\pi\eta r V$. Moving upwards: $qE - mg = 6\pi\eta r (2V) = 2(mg) \implies qE = 3mg$. If field is E/2, upward force is $qE/2 = 1.5mg$. Net upward force = $1.5mg - mg = 0.5mg = 6\pi\eta r V'$. Thus $V' = V/2$.
Question 7:
easyA hollow sphere of radius $1 \text{ m}$ is given a positive charge of $10 \mu \text{C}$. The electric field at the centre of hollow sphere will be: (1990)
The electric field at any point inside a charged hollow conducting sphere is always zero, regardless of the amount of charge on its surface. Therefore, the electric field at the center is zero.