Question

Two parallel plates are charged to produce a potential difference of 45 V. If the separation between the plates is 0.78 m, calculate the magnitude of the electric field in the space between the plates

Answer 1

Answer:

Approximately $58\; {\rm N\cdot C^{-1}}$.

Explanation:

The electric field strength $E$ at a given position is equal to the electric force the field exerts on each unit of electric charge. Hence, the unit of the electric field would be newtons (unit of force) per coulomb (unit of charge.)

Electric field strength is also equal to the change in electric potential over unit distance. For example, in a uniform electric field, if the electric potential changes by $\Delta V$ over a distance of $d$, strength of the electric field would be $E = (\Delta V) / d$.

The electric field between two parallel plates is approximately uniform. Thus, the equation $E = (\Delta V) / d$ can be used to find the strength of that field.

In this question, magnitude of the potential difference between the two charged plates is $\Delta V = 45\; {\rm V}$ over a distance of $d = 0.78\; {\rm m}$. Since this field is uniform, magnitude of the strength of this field would be:

$\begin{aligned}E &= \frac{\Delta V}{d} \\ &= \frac{45\; {\rm V}}{0.78\; {\rm m}} \\ &\approx 58\; {\rm N\cdot C^{-1}}\end{aligned}$.

(Note the unit conversion: $1\; {\rm V} = 1\; {\rm J\cdot C^{-1}} = 1\; {\rm N\cdot m\cdot C^{-1}}$.)

Answer 2

The magnitude of the electric field between the parallel plates is approximately 57.69 V/m. This indicates that for every meter of distance between the plates, there is an electric field strength of 57.69 volts.

The electric field between parallel plates is directly proportional to the potential difference and inversely proportional to the separation distance. This relationship is described by the formula E = V / d, where E is the electric field, V is the potential difference, and d is the separation distance.

The magnitude of the electric field between the parallel plates can be calculated using the formula:

Electric field (E) = Potential difference (V) / Separation distance (d)

Potential difference (V) = 45 V

Separation distance (d) = 0.78 m

Calculating the electric field:

E = V / d

E = 45 V / 0.78 m

E ≈ 57.69 V/m

Therefore, the magnitude of the electric field in the space between the plates is approximately 57.69 V/m.

By substituting the given values into the formula, we can calculate the electric field. In this case, the potential difference is 45 V, and the separation distance is 0.78 m. Dividing the potential difference by the separation distance gives us the magnitude of the electric field.

The magnitude of the electric field between the parallel plates is approximately 57.69 V/m. This indicates that for every meter of distance between the plates, there is an electric field strength of 57.69 volts.

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