The figure a and b show the field lines of a positive and negative point charge, respectively. Give the signs of the potential difference: V P - V Q : V B - V A

From figure a , Electric field lines are drawn from a positive charge.< As electric field lines always flow from higher potential to lower potential, V P > V Q . Thus, V P - V Q is positive. From figure b , Electric field lines are drawn from a negative charge. As electric field lines always flow from higher potential to lower potential, V B > V A . Thus, V B - V A is positive.

The figure a and b show the field lines of a positive and negative point charge, respectively. Give the sign of the potential energy difference of a small negative charge between the points Q and P ; A and B .

Case a : U Q - U P As we know that potential at a point is potential energy per unit charge. ∆ V = ∆ U q where q is the test charge which is taken as a negative charge. So let q = - q o or V Q - V P = U Q - U P - q o or U Q - U P = - q o × V Q - V P . . . i As we know that electric field flows from higher potential to lower potential, V P > V Q . Thus, V Q - V P < 0 . . . ii Using equations i and ii , we get, U Q - U P > 0 , i.e., it is positive. Case b : U A - U B As we know that potential at a point is potential energy per unit charge. ∆ V = ∆ U q where q is the test charge which is taken as a negative charge. So let q = - q o or V A - V B = U A - U B - q o or U A - U B = - q o × V A - V B . . . i As we know that electric field flows from higher potential to lower potential, V B > V A . Thus, V A - V B < 0 . . . ii Using equations i and ii , we get, U A - U B > 0 , i.e., it is positive.

A sample of HCl gas is placed in an electric field of 2 · 5 × 10 - 4 N C - 1 . The dipole moment of each HCl molecule is 3 · 4 × 10 - 30 cm . Find the maximum torque that can act on a molecule.

Given, electric field E = 2 . 5 × 10 - 4 N C - 1 . Dipole moment P = 3 . 4 × 10 - 30 C m . Torque τ = P → × E → = P E sin θ . For maximum torque sin θ must be one. So, τ = PE = 3 . 4 × 10 - 30 × 2 . 5 × 10 4 = 8 . 5 × 10 - 26 N m .

An electric dipole is placed making at an angle 60 ° with an electric field of strength 4 × 10 5 N C - 1 . It experiences a torque equal to 8 3 N - m . Calculate the charge on the dipole, if it is of length 4 cm .

The magnitude of the torque on an electric dipole placed in an electric field is given by τ = P E sin θ . . . i , where P is the dipole moment, E is the electric field and θ is the angle between dipole moment and electric field. Given, θ = 60 ° , E = 4 × 10 5 N C - 1 and τ = 8 3 N - m . Length of dipole 2 a = 4 × 10 - 2 m Putting the values in i , we get, P = 4 × 10 - 5 C - m Using the formula of dipole moment P = q 2 a Putting values of 2 a and P from above, we get, q = 10 - 3 C .