Change on each plate of a parallel plate capacitor is 6microC and capacitance is 4 microf what is p.d between plates?
Ans 1.5 volt
capacitance = charge per unit voltage
4 * 10^-6 = 6 *10^-6 / V
V = 6/4 = 1.5 volts
Thnx
To find the potential difference (p.d.) between the plates of a parallel plate capacitor, we can use the formula:
\( Q = C \cdot V \)
Where:
- \( Q \) is the charge on each plate of the capacitor (6 µC in this case),
- \( C \) is the capacitance of the capacitor (4 µF in this case),
- \( V \) is the potential difference between the plates (what we want to find).
Rearranging the formula to solve for \( V \):
\( V = \frac{Q}{C} \)
Now we can substitute the given values:
\( V = \frac{6 \, \mu C}{4 \, \mu F} \)
As both the numerator and denominator have the same units (µC and µF), we can cancel them out:
\( V = \frac{6}{4} \)
Dividing 6 by 4 gives:
\( V = 1.5 \) (volts)
Therefore, the potential difference between the plates of this parallel plate capacitor is 1.5 volts.