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.

What is an electric dipole? Derive an expression for the electric field intensity at a point on the axis of the dipole.

An electric dipole is a pair of equal and opposite point charges ( q and - q ) that are separated by a small distance( 2 a ). The direction from - q to q is said to be the direction of the dipole. The mid-point of locations of - q and q is called the centre of the dipole. Let point P be at distance r from the centre of the dipole on the side of the charge q as shown in the figure below. Then electric field at P due to the charges of the dipole: E - q = - q 4 π ε o r + a 2 p ^ E + q = - q 4 π ε o r - a 2 p ^ where p ^ is the unit vector along the dipole axis (from - q to + q ). The total electric field at P will be: E → = E → - q + E → + q = q 4 π ε o 1 r - a 2 - 1 r + a 2 p ^ ⇒ E → = q 4 π ε o 4 a r r 2 - a 2 2 p ^ For r > > a E → = 4 q a 4 π ε o r 3 p ^ The above expression is the required expression for the electric field intensity at a point on the axis of the dipole.

Derive Coulomb's law from Gauss' law in electrostatics.

According to Gauss' law, the electric flux through a closed surface S, φ = q ε o where q is the total charge enclosed by the surface. Let us find the electric field at a point P which is at a distance r from a charge q 1 . Draw a spherical gaussian surface that passes through the point P with the charge q 1 at its centre. The electric flux that passes through this surface: φ = E → . A → = q 1 ε o ⇒ E 4 π r 2 cos 0 = q 1 ε o ⇒ E = 1 4 π ε o q 1 r 2 If a charge equal to q 2 is present at the point P, the force experienced by it: F = 1 4 π ε o q 1 q 2 r 2 The above expression gives the force between the two charges according to Coulomb's law.

An electric charge Q is uniformly distributed around a semicircle of radius a . Calculate the electric potential at the centre of this semicircle.

The semicircle can be considered to be made up of charge elements, each equal to d Q . The potential at the centre of the semicircle, due to one such element: d V = 1 4 π ε o d Q a Potential is a scalar quantity hence we can integrate it to get the total potential. The total potential at the centre: V = ∫ 0 V d V = 1 4 π ε o a ∫ 0 Q d Q = 1 4 π ε o Q a The electric potential at the centre of the semicircle is will be Q 4 π ε o a .

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 .