A region in an electric field is specified by the following potential: V x = a x + b Where V is in volt and x is in meter. Find out the units of a and b .

Given, V x = a x + b Units of both sides of the given equation should be the same. Hence unit of b should be volt and the unit of a x should be volt. The unit of x is meter. a x = volt Hence, unit of should a should be voltmetre.

A region in an electric field is specified by the following potential: V x = a x + b Where V is in volt and x is in meter. Find out the units of a and b .

Given, V x = a x + b Units of both sides of the given equation should be the same. Hence unit of b should be volt and the unit of a x should be volt. The unit of x is meter. a x = volt Hence, unit of should a should be voltmetre.

An electric charge + Q is uniformly distributed around a thin metallic circular coil of a single turn and of radius a . Find the expression for the electric potential at a point on its axis at a distance x from its centre.

The circular coil can be considered to be made up of a number of charge elements, each having a value equal to d Q . The distance D of the point on the axis of the coil from any of the charge elements: D = x 2 + a 2 Electric potential due to this charge element: d V = 1 4 π ε o d Q x 2 + a 2 Electric potential is a scalar quantity and charges distributed are all positive hence, the total potential due to all the charge elements distributed on the circular coil will be added up: V = 1 4 π ε o ∫ 0 Q d Q x 2 + a 2 = 1 4 π ε o Q x 2 + a 2 Hence, the expression for the potential at the point on the axis of the charged coil: V = 1 4 π ε o Q x 2 + a 2

A point charge of 3 × 10 - 9 C is released from rest in a uniform electric field and moves a distance of 5 cm after which its kinetic energy becomes 4 . 5 × 10 - 5 J . Calculate the magnitude of the electric field.

Let the mass of the point charge be equal to M kg and the magnitude of the electric field be E N / C . The acceleration produced in the point charge due to electric field: a = 3 × 10 - 9 × E M m / s 2 Using the velocity displacement equation for the constantly accelerated motion of the point charge that starts from the rest, its velocity v after moving through a distance of 5 × 10 - 2 m : v = 2 × 3 × 10 - 9 × 5 × 10 - 2 × E M m / s ...........(i) The kinetic energy: 1 2 M v 2 = 4 . 5 × 10 - 5 J ⇒ v = 4 . 5 × 10 - 5 × 2 M m s - 1 .......(ii) On solving (i) and(ii) we get: E = 3 × 10 5 N / C The magnitude of the electric field is 3 × 10 5 N / C .