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 stat coulomb is placed 2 cm in front of an earthed metallic plane plate of large size. Then the force of attraction on the point charge is

(1) 100 dynes

(2) 160 dynes

(3) 1600 dynes

(4) 400 dynes

Four charges equal to – Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is 

1. $-\frac{Q}{4}(1+2\sqrt{2})$

2. $\frac{Q}{4}(1+2\sqrt{2})$

3. $-\frac{Q}{2}(1+2\sqrt{2})$

4. $\frac{Q}{2}(1+2\sqrt{2})$