Answer:
Part A:
[tex]2Q[/tex]
Part B:
[tex]V_a_b=\frac{Q}{C}[/tex]
Explanation:
The capacitance is given by:
[tex]C=\frac{Q}{V_a_b}[/tex]
Where:
[tex]Q=Stored\hspace{3}electric\hspace{3}charge\\V_a_b=Potential\hspace{3}difference\\C=Capacitance[/tex]
So:
[tex]Q=C V_a_b[/tex]
In the first case:
[tex]V1_a_b=9[/tex]
And in the second case:
[tex]V2_a_b=18[/tex]
or:
[tex]V2_a_b=2*V1_a_b=2*9=18[/tex]
C is a constant so:
[tex]Q_1=C*V1_a_b\\\\so\\V1_a_b=\frac{Q_1}{C} \\ \\\\and\\\\Q_2=C*V2_a_b[/tex]
Therefore:
[tex]Q_2=C*2V1_a_b\\\\Q_2=C*\frac{2Q_1}{C} \\\\Q_2=2*Q_1[/tex]
Part B:
To find thepotential difference in a parallel-plate capacitor just isolate [tex]V_a_b[/tex] from the capacitance equation:
[tex]V_a_b=\frac{Q}{C}[/tex]
Answer:
2Q
Explanation:
The expression for the electric charge stored in a capacitor is given as,
Q = CV..................... Equation 1
Where Q = Electric charge, C = Capacitance of the capacitor, V = Potential difference applied across the plates of the capacitor.
make C the subject of the equation
C = Q/V............... Equation 2
Given: Q = Q, V = 9 V
Substitute into equation 2
C = Q/9 F.
If it is connected to an 18 V battery instead,
Q = CV
Given: C = Q/9 F, V = 18 V
Q = (Q/9)(18)
Q = 2Q.
Hence the charge on the capacitor = 2Q.
