The potential difference between two points x $=$ 10 m and 20 m in the field $E = \frac{100}{x^{2}}$ will be:

5 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $5 V$
$E = - \frac{d V}{d x}$
or $d V = E d x$
integrating, $\int_{V_{10}}^{V_{20}} d V = - \int_{10}^{20} \frac{100}{x^{2}} d x$
or $V_{20} - V_{10} = 100 [\frac{1}{20} - \frac{1}{10}] = - 5$
$∴ V_{10} - V_{20} = 5$ $V$

Suggest Corrections

Similar questions

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

x

E = \frac{100}{x^{2}} N/C

x = 10 m

x = 20 m

D_{1}

D_{2}

D_{3}

- 10 V, - 5 V

- 5 V, - 10 V

- 4 V, - 12 V

60 V

Join BYJU'S Learning Program