Home
Class 12
PHYSICS
Find the potential V of an electrostatic...

Find the potential `V` of an electrostatic field `vec E = a(y hat i + x hat j)`, where `a` is a constant.

A

`v_(0) + ax^(2)y - (ay^(3))/(3)`

B

`V_(0) - axy^(2) - (ay^(2))/(3)`

C

`V_(0) + axy^(2) + (ay^(3))/(3)`

D

`V_(0) - ax^(2)y + (ay^(3))/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • ELECTROSTATICS

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE - IV) (LEVEL - II (ADVANCED) STRAIGHT OBJECTIVE TYPE QUESTIONS)|6 Videos

  • ELECTROSTATICS

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE - IV) (LEVEL - II (ADVANCED) MORE THAN ONE CORRECT ANSWER TYPE QUESTIONS)|4 Videos

  • ELECTROSTATICS

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE - III) (LEVEL - II (ADVANCED) INTEGER TYPE QUESTIONS)|2 Videos

  • ELECTROMAGNETICS

    AAKASH SERIES|Exercise ADDITIONAL EXERCISE|13 Videos

  • ELEMENTS OF VECTORS

    AAKASH SERIES|Exercise QUESTIONS FOR DESCRIPTIVE ANSWERS|10 Videos

Similar Questions

Explore conceptually related problems

The potential field of an electric field vec(E)=(y hat(i)+x hat(j)) is

The potential field of an electric field vec(E)=(y hat(i)+x hat(j)) is

A particle of charge q and mass m starts moving from the origin under the action of an electric field vec E = E_0 hat i and vec B = B_0 hat i with a velocity vec v = v_0 hat j . The speed of the particle will become 2v_0 after a time.

A particle of charge q and mass m starts moving from the origin under the action of an electric field vec E = E_0 hat i and vec B = B_0 hat i with a velocity vec v = v_0 hat j . The speed of the particle will become 2v_0 after a time.

A particle of charge q and mass m starts moving from the origin under the action of an electric field vec E = E_0 hat i and vec B = B_0 hat i with a velocity vec v = v_0 hat j . The speed of the particle will become 2v_0 after a time.

Charge Q is given a displacement vec r = a hat i + b hat j in an electric field vec E = E_1 hat i + E_2 hat j . The work done is.

Find V_(ab) in an electric fiels vec E = (2 hat I + 3 hat j + 4 hat k) NC^-1 where vec r_a = (hat I - 2 hat j + hat k) m and vec r_b = (2 hat i + hat j - 2 hat k) m .

Potential difference between two points (V_(A) - V_(B)) in an electric field E = (2 hat(i) - 4 hat(j)) N//C , where A = (0, 0) and B = (3m, 4m) is

Find the potential difference V_(AB) between A (0,0,0) and B (1 m , 1 m,1 m) in an electric field : (i) vec E = (y hat i + x hat j) Vm^-1 (ii) vec E = (3 x^2 y hat i + x^3 hat j) Vm^-1 .

A charged particle moves with velocity vec v = a hat i + d hat j in a magnetic field vec B = A hat i + D hat j. The force acting on the particle has magnitude F. Then,