The charge on 500 cc of water due to protons will be:

1. 6.0 × 10²⁷ C
2. 2.67 × 10⁷ C
3. 6 × 10²³ C
4. 1.67 × 10²³ C

In the given figure two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal length L. Assume that θ is so small that $\tan \theta \approx \sin \theta$, then for equilibrium x is equal to 

(1) ${\left(\frac{{\displaystyle q^{2}L}}{{\displaystyle 2\pi {\epsilon }_{0}mg}}\right)}^{\frac{1}{3}}$

(2) ${\left(\frac{{\displaystyle q{L}^2}}{{\displaystyle 2\pi {\epsilon }_{0}mg}}\right)}^{\frac{1}{3}}$

(3) ${\left(\frac{{\displaystyle q^2{L}^2}}{{\displaystyle 4\pi {\epsilon }_{0}mg}}\right)}^{\frac{1}{3}}$

(4) ${\left(\frac{{\displaystyle q^{2}L}}{{\displaystyle 4\pi {\epsilon }_{0}mg}}\right)}^{\frac{1}{3}}$

Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in 

(1)   (2)

(3)    (4)

Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when 

(1) $x=\frac{d}{\sqrt{2}}$

(2) $x=\frac{d}{2}$

(3) $x=\frac{d}{2\sqrt{2}}$

(4) $x=\frac{d}{2\sqrt{3}}$

An electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertia is I. If the dipole is displaced slightly from the equilibrium position, then the angular frequency of its oscillations is?

1. ${\left(\frac{{\displaystyle pE}}{{\displaystyle I}}\right)}^{1/2}$

2. ${\left(\frac{{\displaystyle pE}}{{\displaystyle I}}\right)}^{3/2}$

3. ${\left(\frac{{\displaystyle I}}{{\displaystyle pE}}\right)}^{1/2}$

4. ${\left(\frac{{\displaystyle p}}{{\displaystyle IE}}\right)}^{1/2}$

An infinite number of electric charges each equal to $5$ nC (magnitude) are placed along the $x\text-$axis at $x=1$ cm, $x=2$ cm, $x=4$ cm, $x=8$ cm ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at $x=0$ is: $\left(\frac{1}{4 \pi \varepsilon_{0}} = 9 \times10^{9} ~\text{N-m}^{2}/\text{C}^{2}\right)$
1. $12\times 10^{4}$
2. $24\times 10^{4}$
3. $36\times 10^{4}$
4. $48\times 10^{4}$

Three charges –q₁, +q₂ and –q₃ are placed as shown in the figure. The x-component of the force on –q₁ is proportional to 

(1) $\frac{q_2}{b^2}-\frac{q_3}{a^2}\sin \theta$

(2) $\frac{q_2}{b^2}-\frac{q_3}{a^2}\cos \theta$

(3) $\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta$

(4) $\frac{q_2}{b^2}+\frac{q_3}{a^2}\cos \theta$

Two-point charges $+q$ and $–q$ are held fixed at $(–d, 0)$ and $(d, 0)$ respectively of a $(x, y)$ coordinate system. Then:

A point charge of $40~\text{stat coulomb }$is placed $2~\text{cm}$ in front of an earthed metallic plane plate of large size. Then the force of attraction on the point charge is
1. $100~\text{dynes}$
2. $160~\text{dynes}$
3. $1600~\text{dynes}$
4. $400~\text{dynes}$