1. Given data

• Specific charge of body: $\frac{q}{m}=8\,\mu\text{C/g}$
• Distance from wall: $d=10\text{ cm}=0.1\text{ m}$
• Electric field: $E=100\text{ V/m}=100\text{ N/C}$
• Plane is frictionless.
• Collision with wall is perfectly elastic.

2. Convert specific charge into SI units

We have: $8\,\mu\text{C/g}=8\times 10^{-6}\text{ C}/10^{-3}\text{ kg}$

So, $\frac{q}{m}=8\times 10^{-3}\text{ C/kg}$

3. Find acceleration of the body

Electric force on the charged body: $F=qE$

Hence acceleration: $a=\frac{F}{m}=\frac{qE}{m}=\left(\frac{q}{m}\right)E$

Substitute values: $a=(8\times 10^{-3})(100)=0.8\text{ m/s}^2$

4. Time taken to reach the wall

Initially the body is at rest and moves distance $d=0.1$ m under constant acceleration $a=0.8$ m/s$^2$.

Using $d=\frac{1}{2}at^2$

So, $0.1=\frac{1}{2}(0.8)t^2$ $0.1=0.4t^2$ $t^2=\frac{0.1}{0.4}=0.25$ $t=0.5\text{ s}$

This is the time to go from the starting point to the wall.

5. Motion after elastic collision

At the wall, the body collides elastically, so its velocity reverses instantly.

The electric field still acts toward the wall, so acceleration remains toward the wall. Thus after collision, the body moves away from the wall, slows down under the electric force, comes to rest at the original position, and then repeats the motion.

Therefore, one complete cycle consists of:

• going from initial position to wall: $0.5$ s
• returning from wall to initial position: $0.5$ s

Hence time period: $T=0.5+0.5=1\text{ s}$

6. Final answer

$\boxed{1}$