Question 325: Electrostats & Capacitance

Question

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V . The potential difference between the points A and $B$ (shown in the figure) is :

Accepted Answer

Step 1: Formula for Potential Difference
The potential difference ($$\Delta V$$) between two points in a uniform electric field is found by multiplying the electric field ($$E$$) by the distance ($$\Delta d$$) between the points:

$$\Delta V = E 	imes \Delta d$$
Step 2: Find the Electric Field
The total potential difference between the two plates is $$V$$, and they are 10 cm apart.

So, the electric field between the plates is:

$$E = \frac{V}{10}$$

Here, distance is in centimeters.
Step 3: Calculate Potential Difference Between A and B
Points A and B are 4 cm apart.

So, the potential difference between A and B will be:

$$V_{AB} = E 	imes 4 = \frac{V}{10} 	imes 4 = \frac{4V}{10} = \frac{2V}{5}$$

Answer

1 V

Answer

$\frac{1}{4} \mathrm{~V}$

Answer

$\frac{3}{4} \mathrm{~V}$

Answer

$\frac{2}{5} \mathrm{~V}$

Electrostats & Capacitance

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V . The potential difference between the points A and B (shown in the figure) is :

Choose the correct answer: