Answer:
The potential difference across the plates is 226 V.
Explanation:
Given;
area of the capacitor plate, A = 0.2 m²
separation, d = 0.1 mm = 0.1 x 10⁻³ m
charge on each plate, Q = 4 x 10⁻⁶ C
Charge on the capacitor is given by;
Q = CV
Where;
C is the capacitance of the capacitor, given as;
C = ε₀A / d
Then, the potential difference across the plates is given by;

Therefore, the potential difference across the plates is 226 V.
<h3>No:1</h3>
The object is moving with constant or uniform acceleration and in average speed
<h3>No:-2</h3>
The object is de accelerating
<h3>No:-3</h3>
The object deaccelerated and came to rest so fast.
<h3>No:-4</h3>
The object moves slowly first then accelerated.
<h3>No:-5</h3>
The object accelerated at first so fast then move with constant acceleration then again accelerated .