Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
Explanation:
Relation between potential energy and charge is as follows.
U = qV
or, 
= 
=
J
or, = 
Therefore, we can conclude that change in the electrical potential energy
is
.
<h3>Option B</h3><h3>The time constant of a 10 H inductor and a 200 ohm resistor connected in series is 50 millisecond</h3>
<em><u>Solution:</u></em>
Given that,
10 H inductor and a 200 ohm resistor connected in series
To find: time constant
<em><u>The time constant in seconds is given as:</u></em>

Where,
L is the inductance in henry and R is the resistance in ohms

Convert to millisecond
1 second = 1000 millisecond
0.05 second = 0.05 x 1000 = 50 millisecond
Thus time constant is 50 millisecond