Home
Class 12
PHYSICS
A tiny electric dipole of dipole moment ...

A tiny electric dipole of dipole moment `vecP = P_0hatj` is placed at point (l,0). There exists an electric field `vecE = 2ax^2hati + (2by^2 + 2cy)hatj`.

A

Force on dipole is `2P_0ahati`.

B

Force on dipole is `2P_0bhati`.

C

Force on dipole is `2P_0chatj`.

D

Force on dipole is `-2P_0chatj`.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the force acting on the electric dipole placed in the given electric field. The dipole moment is given as \(\vec{P} = P_0 \hat{j}\) and the electric field is given as \(\vec{E} = 2a x^2 \hat{i} + (2b y^2 + 2cy) \hat{j}\). The dipole is placed at the point \((l, 0)\). ### Step 1: Identify the components of the dipole moment and electric field The dipole moment \(\vec{P}\) has components: - \(P_x = 0\) (since it only has a \(\hat{j}\) component) - \(P_y = P_0\) The electric field \(\vec{E}\) has components: - \(E_x = 2a x^2\) - \(E_y = 2b y^2 + 2cy\) ### Step 2: Calculate the force on the dipole The force \(\vec{F}\) on a dipole in an electric field is given by: \[ \vec{F} = P_x \frac{\partial E_x}{\partial x} + P_y \frac{\partial E_y}{\partial y} \] ### Step 3: Evaluate the force in the x-direction Since \(P_x = 0\): \[ F_x = P_x \frac{\partial E_x}{\partial x} = 0 \] ### Step 4: Evaluate the force in the y-direction Now we need to calculate \(F_y\): \[ F_y = P_y \frac{\partial E_y}{\partial y} \] Substituting \(P_y = P_0\): \[ F_y = P_0 \frac{\partial E_y}{\partial y} \] ### Step 5: Differentiate \(E_y\) with respect to \(y\) The expression for \(E_y\) is: \[ E_y = 2b y^2 + 2cy \] Differentiating with respect to \(y\): \[ \frac{\partial E_y}{\partial y} = \frac{\partial}{\partial y}(2b y^2 + 2cy) = 4by + 2c \] ### Step 6: Substitute the coordinates of the dipole The dipole is located at the point \((l, 0)\), so we substitute \(y = 0\): \[ \frac{\partial E_y}{\partial y} \bigg|_{y=0} = 4b(0) + 2c = 2c \] ### Step 7: Calculate \(F_y\) Now substituting back into the expression for \(F_y\): \[ F_y = P_0 (2c) = 2P_0 c \] ### Step 8: Write the final force vector Since \(F_x = 0\) and \(F_y = 2P_0 c\), we can write the force vector as: \[ \vec{F} = 0 \hat{i} + 2P_0 c \hat{j} = 2P_0 c \hat{j} \] ### Final Answer The force on the dipole is: \[ \vec{F} = 2P_0 c \hat{j} \] ---
To solve the problem, we need to find the force acting on the electric dipole placed in the given electric field. The dipole moment is given as \(\vec{P} = P_0 \hat{j}\) and the electric field is given as \(\vec{E} = 2a x^2 \hat{i} + (2b y^2 + 2cy) \hat{j}\). The dipole is placed at the point \((l, 0)\). ### Step 1: Identify the components of the dipole moment and electric field The dipole moment \(\vec{P}\) has components: - \(P_x = 0\) (since it only has a \(\hat{j}\) component) - \(P_y = P_0\) The electric field \(\vec{E}\) has components: ...
Promotional Banner

Topper's Solved these Questions

  • MISCELLANEOUS VOLUME 3

    CENGAGE PHYSICS ENGLISH|Exercise Assertion and Reason Type|8 Videos

  • MISCELLANEOUS VOLUME 3

    CENGAGE PHYSICS ENGLISH|Exercise Comprehension Type|94 Videos

  • MAGNETIC FIELD AND MAGNETIC FORCES

    CENGAGE PHYSICS ENGLISH|Exercise Multiple Correct Answer type|2 Videos

  • MISCELLANEOUS VOLUME 5

    CENGAGE PHYSICS ENGLISH|Exercise Integer|12 Videos

Similar Questions

Explore conceptually related problems

Electric dipole of moment vecp = phati is kept at a point (x,y) in an electric field vecE = 4xy^2hati + 4x^2yhatj . Find the force on the dipole.

An electric dipole of dipole moment p is placed in a uniform external electric field E. Then, the

Electric dipole of moment vec(P)=P hat(i) is kept at a point (x, y) in an electric field vec(E )=4x y^(2)hat(i)+4x^(2)y hat(j) . Find the magnitude of force acting on the dipole.

A dipole of dipole moment vec(P) is placed in uniform electric field vec(E) , then torque acting on it is given by

A dipole of dipole moment vecp=phati lies along the x -axis in a non-uniform electric field vecE=(c)/(x)hati . The force acting on the dipole is

An electric dipole of dipole moment vecP is placed in a uniform electric field vecE such that vecP is perpendicular to vecE . The work done to turn the dipole through an angle of 180^(@) is -

An electric dipole of dipole moment vecp is placed in a uniform electric field vecE with its axis inclined to the field. Write an expression for the torque vectau experienced by the dipole in vector form. Show diagrammatically how the dipole should be kept in the electric field so that the torque acting on it is: (i) maximum (ii) zero

An electric dipole of dipole moment vecp is placed in a uniform electric field vecE with its axis inclined to the field. Write an expression for the torquet vec tau experienced by the dipole in vector form. Show diagrammatically how the dipole should be kept in the electric field so that the torque acting on it is : maximum

An electric dipole of dipole moment vecp is placed in a uniform electric field vecE with its axis inclined to the field. Write an expression for the torquet vec tau experienced by the dipole in vector form. Show diagrammatically how the dipole should be kept in the electric field so that the torque acting on it is : zero

An electric dipole of dipole moment p = 6 * 10^-4 C m is placed in a uniform external electric field of intensity 2*10^5(N/C) as shown in the figure, The anglebetween net electric field with external electric field at point A will be (epsilon