Q14PE - College Physics (Urone)

Question

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }\mathbf{V}$?

Step-by-Step Solution

Answer

 $1.50\times {10}^6 \mathrm{V}/\mathrm{m}$ is the required electric field strength.

Step 1: Principle

The magnitude of the electric field (E) at any point is given by the potential gradient at that point.

 $\mathbf{E}\mathbf{=}\left|-\frac{\mathrm{dV}}{\mathrm{dr}}\right|$

Here dV is the potential gradient and dr is the displacement from source charge.

Step 2: The given data

• The distance between the plates is: 

           $\begin{array}{rcl}\mathrm{d}& =& (1.00 \mathrm{cm})\left(\frac{1 \mathrm{m}}{100 \mathrm{cm}}\right)\\ & =& 0.0100 \mathrm{m}\\ & & \end{array}$

• The potential difference between the two plates is: $\mathrm{\Delta V}=1.50\times {10}^4 \mathrm{V}.$  

Step 3: Calculation of electric field strength

Equation (1) is used to calculate the electric field strength E between the two plates:

 $\mathbf{E}\mathbf{=}\frac{\mathbf{\Delta V}}{\mathbf{d}}$

 

After entering the numbers for $∆\mathrm{V}$ and d,

 $\begin{array}{rcl}\mathbf{E}& \mathbf{=}& \frac{\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }\mathbf{V}}{\mathbf{0}\mathbf{.}\mathbf{0100}\mathbf{ }\mathbf{m}}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{V}\mathbf{/}\mathbf{m}\\ & & \end{array}$

Therefore, the electric field strength is $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{V}\mathbf{/}\mathbf{m}$.