Electric Potential: Practice Problem & Solution

Electric field at a distance x from origin is given as E = 100/x², then potential difference between points situated at x = 10 m and x = 20 m.

5 V

10 V

15 V

4 V

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:

Using

\Delta V = -\int_{x_1}^{x_2} E \, dx

for

E = \frac{100}{x^2}

\Delta V = -\int_{10}^{20} \frac{100}{x^2} \, dx = \left[ \frac{100}{x} \right]_{10}^{20} = \frac{100}{20} - \frac{100}{10} = -5\text{ V}

The magnitude of the potential difference is $5\text{ V}$ .

Detailed Solution (Best of Understanding Complex Concepts)

Given Data

Electric field function: $E(x) = \frac{100}{x^2}\text{ N/C}$
Initial position ( $x_1$ ): $10\text{ m}$
Final position ( $x_2$ ): $20\text{ m}$

Step 1: Relation Between Electric Field and Potential

The potential difference

\Delta V

between two points

x_1

and

x_2

in an electric field

E(x)

is defined by the line integral:

\Delta V = V(x_2) - V(x_1) = -\int_{x_1}^{x_2} E(x) \, dx

Step 2: Set Up the Definite Integral

Substitute the given values

x_1 = 10\text{ m}

x_2 = 20\text{ m}

, and

E(x) = \frac{100}{x^2}

into the formula:

\Delta V = -\int_{10}^{20} \frac{100}{x^2} \, dx

Factor out the constant

100

\Delta V = -100 \int_{10}^{20} x^{-2} \, dx

Step 3: Integrate the Function

Using the standard integration rule

\int x^n \, dx = \frac{x^{n+1}}{n+1}

\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x}

Now, substitute this back into the expression:

\Delta V = -100 \left[ -\frac{1}{x} \right]_{10}^{20}

The negative sign inside and outside the bracket cancel each other out:

\Delta V = \left[ \frac{100}{x} \right]_{10}^{20}

Step 4: Evaluate the Limits

Apply the fundamental theorem of calculus

\left[ f(x) \right]_a^b = f(b) - f(a)

\Delta V = \left( \frac{100}{20} - \frac{100}{10} \right)

\Delta V = 5 - 10 = -5\text{ V}

Conclusion

The potential difference $V(20) - V(10)$ is $-5\text{ V}$ .
The magnitude of the potential difference $\vert{}\Delta V\vert{}$ is $5\text{ V}$ .