Question

Two plates of area 30.0 cm^2 are separated by a distance of 0.0590 cm. If a charge separation of 0.0240 μC is placed on the two plates, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.

Answer 1

The potential difference between the two plates is 4.425 volts.

Step-by-step explanation:

To calculate the potential difference (voltage) between the two plates, we can use the formula: V = Ed

where V is the potential difference, E is the electric field strength, and d is the separation distance between the plates.

Given that the electric field strength is 7.50 x 10^4 V/m and the separation distance is 0.0590 cm (or 0.0000590 m), we can plug in these values to calculate the potential difference: V = (7.50 x 10^4 V/m) × (0.0000590 m)

Simplifying the expression gives us the potential difference between the plates: V = 4.425 V

Therefore, the potential difference between the two plates is 4.425 volts.

Answer 2

V=533.33 V

Step-by-step explanation:

Given that

A= 30 cm²

d= 0.059 cm

Q= 0.0240 μ C

We know that capacitance given as

`C=(\varepsilon _oA)/(d)`

Now by putting the values

`C=(8.85* 10^(-12)* 30* 10^(-4))/(0.059* 10^(-2))`

`C=4.5* 10^(-11)\ F`

Voltage difference given as

Q= V .C

V=Q/C

`V=\frac{0.0240* 10^(-6)} {4.5* 10^(-11)}`

V=533.33 V