Answer:
Explanation:
Given a parallel plate capacitor of
Area=A
Distance apart =d
Potential difference, =V
If the distance is reduce to d/2
What is p.d
We know that
Q=CV
Then,
V=Q/C
Then this shows that the voltage is inversely proportional to the capacitance
Therefore,
V∝1/C
So, VC=K
Now, the capacitance of a parallel plate capacitor is given as
C= εA/d
When the distance apart is d
Then,
C1=εA/d
When the distance is half d/2
C2= εA/(d/2)
C2= 2εA/d
Then, applying
VC=K
V1 is voltage of the full capacitor V1=V
V2 is the required voltage let say V'
Then,
V1C1=V2C2
V × εA/d=V' × 2εA/d
VεA/d = 2V'εA/d
Then the εA/d cancels on both sides and remains
V=2V'
Then, V'=V/2
The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2
Answer:
The potential difference reduced by half of its value 0.5V
Explanation:
Potential difference
Potential difference is the difference in the potential energy of two points along a conductor carry a charge.
Capacitance
Capacitance is the measure of energy that is stored in a capacitor which is a circuit component that stores energy
Step by Step Calculation
The capacitance, Charge and potential relationship can be expressed in equation 1
Q = CV .....................1
where Q is the charge
C is the Capacitance and
V is the potential difference
but C = εoA/d ....................2
Substitution the value of C in equation 2 into equation 1 we have
Q = εoA/d x V
making the potential difference the subject formula we have
V = Qd/εoA ...........................3
since the separation was decreased by d/2 we have a new value for d and is substituted into equation 3 and V is given a new value as [tex]V_{1\\}[/tex]
[tex]V_{1\\}[/tex] = [tex]\frac{Qd}{2e_{o}A }[/tex]
To obtain the final potential difference we compute the ratio of V and [tex]V_{1\\}[/tex]
V:[tex]V_{1\\}[/tex] = Qd/εoA : Qd/2εoA
V:[tex]V_{1\\}[/tex] = 1:2
[tex]V_{1\\}[/tex] = 0.5 V
So the new potential difference is 0.5 V which halves of the first potential difference.
