The potential function of an electrostatic field is given by $V = 2 x^{2}$ . Determine the electric field strength at the point $(2 m, 0, 3 m)$ :

¯ E = 4^i (N C^{- 1})

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

¯ E = - 4^i (N C^{- 1})

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

¯ E = 8^i (N C^{- 1})

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

¯ E = - 8^i (N C^{- 1})

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

Open in App

Solution

The correct option is B $¯ E = - 8^i (N C^{- 1})$

Electric field in x direction $E_{x} = - \frac{d v}{d x} = - 4 x = - 4 \times 2 = - 8$

Electric field in y and z direction $E_{y} = - \frac{d v}{d y} = 0, E_{z} = - \frac{d v}{d z} = 0$

Hence $¯ E = - 8^i N C^{- 1}$

Suggest Corrections

Similar questions

V = 2 x^{2} .

(2 m, 0, 3 m) .

The potential at the origin is zero due to electric field $¯ E = 20^i + 30^j N C^{- 1}$ . The potential at point $P (2 m, 2 m)$ is:

V

¯ E = a (y^i + x^j)

V = 2 x^{2} .

(2 m, 0, 3 m) .

¯ E a n d ¯ B

¯ X

¯ k

Join BYJU'S Learning Program