1. Given data

• Mass: $m=1\,\text{kg}$
• Charge: $q=5\,\text{mC}=5\times 10^{-3}\,\text{C}$
• Electric field: $E=200\,\text{N/C}$ (horizontal)
• Inclination: $\theta=30^\circ$
• Coefficient of friction: $\mu=0.2$
• Vertical height: $h=1\,\text{m}$
• Initial velocity: $u=0$

Also, $\sin 30^\circ=\frac12, \qquad \cos 30^\circ=\frac{\sqrt 3}{2}$

2. Length of the incline

If the body starts from a vertical height $h=1\,\text{m}$, then the distance along the plane is $s=\frac{h}{\sin 30^\circ}=\frac{1}{1/2}=2\,\text{m}$

3. Forces along and perpendicular to the plane

(a) Gravitational force

Component of weight along the plane: $mg\sin\theta=1\times 9.8\times \frac12=4.9\,\text{N}$

Component perpendicular to the plane: $mg\cos\theta=1\times 9.8\times \frac{\sqrt 3}{2}\approx 8.49\,\text{N}$

(b) Electric force

Magnitude of electric force: $F_e=qE=(5\times 10^{-3})(200)=1\,\text{N}$

Since the electric field is horizontal and the plane is inclined at $30^\circ$ to the horizontal:

• Component of electric force along the plane: $F_{e,\parallel}=F_e\cos 30^\circ=1\cdot \frac{\sqrt 3}{2}\approx 0.866\,\text{N}$ This acts up the plane if the field is horizontal toward the right as in the usual figure while the block slides down the plane.

• Component of electric force perpendicular to the plane: $F_{e,\perp}=F_e\sin 30^\circ=1\cdot \frac12=0.5\,\text{N}$ This acts into the plane, increasing the normal reaction.

4. Normal reaction

Thus, $N=mg\cos\theta+F_{e,\perp}=8.49+0.5=8.99\,\text{N}$

5. Friction force

Since the block moves down the plane, friction acts up the plane.

$f=\mu N=0.2\times 8.99=1.798\,\text{N}$

6. Net force down the plane

Taking downward along the plane as positive, $F_{\text{net}}=mg\sin\theta-F_{e,\parallel}-f$ $F_{\text{net}}=4.9-0.866-1.798=2.236\,\text{N}$

So acceleration is $a=\frac{F_{\text{net}}}{m}=\frac{2.236}{1}=2.236\,\text{m/s}^2$

7. Use equation of motion

The block starts from rest and travels distance $s=2\,\text{m}$: $s=ut+\frac12 at^2$ $2=0+\frac12(2.236)t^2$ $2=1.118\,t^2$ $t^2=\frac{2}{1.118}\approx 1.789$ $t\approx \sqrt{1.789}\approx 1.34\,\text{s}$

Thus, $t\approx 1.3\,\text{s}$

8. Option check

• A: $0.46\,\text{s}$ ✗
• B: $0.92\,\text{s}$ ✗
• C: $1.3\,\text{s}$ ✓
• D: $2.3\,\text{s}$ ✗

So the correct option is C.