Answer:
3 V
Explanation:
The relationship between capacity (C), charge stored (Q) and potential difference across a capacitor (V) is
[tex]C=\frac{Q}{V}[/tex] (1)
The problem asks us to find the emf of the battery several seconds after the switch has been closed: this means that the capacitor had enough time to fully charge, so the potential difference across the capacitor (V) is equal to the emf of the battery.
Since we have:
[tex]C=10 \mu F=10\cdot 10^{-6} F[/tex]
[tex]Q=30 \mu C=30\cdot 10^{-6} C[/tex]
We can solve the equation for V, and we find
[tex]V=\frac{Q}{C}=\frac{30 \cdot 10^{-6}C}{10 \cdot 10^{-6} F}=3 V[/tex]
A switch that connects a battery to a 10 µF capacitor is closed for several seconds and the capacitor plates are charged to 30 µC then the voltage across the capacitor will be 3 V.
The emf of the battery will be equal to the voltage across the capacitor which is 3 V.
The EMF or electromotive force can be defined as the energy supplied by a battery or a cell per coulomb (Q) of charge passing through it. The magnitude of emf is equal to V (potential difference) across the cell terminals when there is no current flowing through the circuit.
Hence the emf of the battery will be equal to the voltage across the capacitor.
The voltage across a capacitor can be calculated as given below.
[tex]V = \dfrac {Q}{C}[/tex]
Where Q is the amount of charge stored on each plate and C is the capacitance.
[tex]V = \dfrac { 30 }{10}[/tex]
[tex]V = 3\;\rm V[/tex]
Hence we can conclude that the emf of the battery is 3 V.
To know more about the emf and voltage, follow the link given below.
https://brainly.com/question/118936.
