A charge particle of charge + q , mass m and moving with a velocity of u i ^ enters a uniform electric field of strength E → = E j ^ . Find magnitude of velocity and magnitude of displacement of the particle after time t .

Given that, Uniform electric field of strength is given as E → = E j ^ Acceleration of charged particle of mass m will be a = q E m j ^ This implies that a x = 0 and a y = q E m After time t , s x = v x t ⇒ s x = ( u × t ) Similarly, s y = u y t + 1 2 a y t 2 s y = 0 + 1 2 × q E m × t 2 s y = q E t 2 2 m Therefore, net displacement , s = s x i ^ + s y j ^ ⇒ s = u t i ^ + q E t 2 2 m j ^ Also, the particle is moving with velocity u i ^ This implies that v x = u (constant as a x = 0 ) & v y = u y + a y t ⇒ v y = 0 + q E m t ⇒ v y = q E t m So, net magnitude of velocity will be v = v x i ^ + v y j ^ = u i ^ + q E t m j ^

Show that the electric field at the surface of a charged conducting sphere is given by E → = σ ε 0 n ^ , where σ is the surface charge density and n ^ is a unit vector normal to the surface in the outward direction.

Let a charge Q be given to a conductor, this charge under electrostatic equilibrium will redistribute and the electric field inside the conductor is zero (i.e., E i n = 0 ). Let us consider a point P at which electric field strength is to be calculated, just outside the surface of the conductor. Let the surface charge density on the surface of the conductor in the neighbourhood of P be σ . Now consider a small cylindrical box CD having one base C passing through P; the other base D lying inside the conductor and the curved surface being perpendicular to the surface of the conductor. Let the area of each flat base be a . As the surface of the conductor is equipotential surface, the electric field strength E at P, just outside the surface of the conductor is perpendicular to the surface of the conductor in the neighbourhood of P. The flux of electric field through the curved surface of the box is zero, since there is no component of electric field E normal to curved surface. Also, the flux of electric field through the base D is zero, as electric field strength inside the conductor is zero. Therefore, the resultant flux of electric field through the entire surface of the box is same as the flux through the face C. This may be analytically seen as: If S 1 and S 2 are flat surfaces at C and D and S 3 is curved surface, then, Total electric flux, ∫ S E → . d S → = ∫ S 1 E → . d S 1 → + ∫ S 2 E → . d S → 2 + ∫ S 3 E → . d S → 3 = ∫ S 1 E . d S 1 cos 0 ∘ + ∫ S 2 0 . d S → 2 + ∫ S 3 E . d S 3 cos 90 ∘ = ∫ S 1 E . d S 1 = E a As the charge enclosed by the cylinder is ( σ a ) coulomb, we have, using Gauss’s theorem, Total electric flux = 1 ∈ ∘ × Charge enclosed Thus, the electric field strength at any point close to the surface of a charged conductor of any shape is equal to 1 ∈ ∘ times the surface charge density σ . This is known as Coulomb’s law. The electric field strength is directed radially away from the conductor if σ is positive and towards the conductor if S is negative. If n ^ is unit vector normal to surface in outward direction, then E → = σ ∈ ∘ n ^ Obviously electric field strength near a plane conductor is twice of the electric field strength near a non-conducting thin sheet of charge.

(i) Express Coulomb's law in vector form. (ii) Two point charges q 1 = 5 × 10 - 6 C and q 2 = 3 × 10 - 6 C are located at ( 3 , 5 , 1 ) m and ( 1 , 3 , 2 ) m . Find F 12 → and F 21 → using vector form of Coulomb's law.

(i) The Coulomb’s law of electrostatic force states that the electrostatic force of attraction and repulsion are directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. ⇒ F ∝ q 1 × q 2 ⇒ F ∝ 1 r 2 Where, q 1 , q 2 are the charges and r is the distance between the charges Thus, the electrostatic force is given by, ⇒ F = k q 1 q 2 r 2 Where, k is the constant of proportionality The vector form of the Coulomb’s law is given as: F → = k q 1 q 2 r 2 12 r ^ 12 Here, r 12 is the displacement from charge 1 to charge 2. The magnitude of force between charge 1 and charge 2 is equal. But in vector form force on charge 1 is given as: F → 12 = k q 1 q 2 r 2 12 r ^ 12 Force on charge 2 is given as: F → 21 = k q 1 q 2 r 2 21 r ^ 21 ⇒ F → 12 = F → 21 (ii) Here, q 1 = 5 X 10 - 6 C q 2 = 3 X 10 - 6 C r → 2 = ( 3 i ^ + 5 j ^ + k ^ ) r → 1 = ( i ^ + 2 j ^ + 2 k ) ^ r → 12 = r → 2 − r → 1 = ( 2 i ^ + 2 j ^ − k ^ ) r → 12 = ( 2 ) 2 + ( 2 ) 2 + ( - 1 ) 2 = 3 m r ^ 12 = r → 12 r → 12 = 2 i ^ + 2 j ^ - k ^ 3 F → 21 = k q 1 q 2 r 2 12 r ^ 12 = 9 × 10 9 × 5 × 10 - 6 × 3 × 10 - 6 ( 3 ) 2 × 2 i ^ + 2 j ^ - k ^ 3 = 5 × 10 - 3 × ( 2 i ^ + 2 j ^ - k ^ ) N F → 21 = 2 2 + 2 2 + - 1 2 × 5 × 10 - 3 = 1 . 5 × 10 - 2 N As F → 12 = F → 21 = - ( 5 × 10 - 3 ) × ( 2 i ^ + 2 j ^ - k ^ ) N

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A charged wire $A B$ of length $l$ is placed inside a sphere, as shown in figure. Linear charge density of wire is $λ = k x$ , where $x$ is the distance measured along the wire from end $A$ and $K$ is constant. Calculate the flux through the sphere.

Important Questions on Electric Charges and Fields

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 38

Two plane sheets of charge densities $+ σ$ and $- σ$ are kept in air, as shown in figure. What are the electric field intensities at points $A$ and $B$ ?

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 39

A charge particle of charge

+ q

, mass

m

and moving with a velocity of

u \hat{i}

enters a uniform electric field of strength

\vec{E} = E \hat{j}

. Find magnitude of velocity and magnitude of displacement of the particle after time

t

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 40

A uniformly charged conducting sphere of

2.5 m

in diameter has a surface charge density of

100 μ C m^{- 2}

. Calculate the
(i) Charge on the sphere.
(ii) Total electric flux coming out a Gaussian surface just enclosing the outer surface of the sphere.

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 42

A spherical conducing shell of inner radius

r_{1}

and outer radius

r_{2}

has a charge

Q

. Another charge

q

is placed at the centre of the shell.
(a) What is the surface charge density on the
(i) Inner surface?
(ii) Outer surface of the shell?
(b) Derive the expression for the electric field at a point

x > r_{2}

from the centre of the shell.

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 43

Show that the electric field at the surface of a charged conducting sphere is given by

\vec{E} = \frac{σ}{ε_{0}} \hat{n}

, where

σ

is the surface charge density and

\hat{n}

is a unit vector normal to the surface in the outward direction.

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 44

Two small identical electric dipoles $A B$ and $A C$ , each of dipole moment $p$ are kept at an angle of $120 °$ as shown in figure. What is the resultant dipole moment of this combination? If this system is subjected to electric field $(\vec{E})$ directed along $+ x$ direction, what will be the magnitude and direction of torque acting on this?

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 46

(i) State Coulomb's law and write the formula of electrostatic force between two charges, separated by certain distance. Under which condition this formula is applicable?
(ii) Two positive point charges which are

0.1 m

apart repel each other with a force of

18 N

. If the sum of the charges be

9 μ C

, calculate their separate values of charges.

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Gamma Question Bank for Medical Physics>Chapter 16 - Electric Charges and Fields>Assignment>Q 47

(i) Express Coulomb's law in vector form.
(ii) Two point charges

q_{1} = 5 \times 10^{- 6} C

and

q_{2} = 3 \times 10^{- 6} C

are located at

(3, 5, 1) m

and

(1, 3, 2) m

. Find

\vec{F_{12}}

and

\vec{F_{21}}

using vector form of Coulomb's law.

A charged wire AB of length l is placed inside a sphere, as shown in figure. Linear charge density of wire is λ=kx, where x is the distance measured along the wire from end A and K is constant. Calculate the flux through the sphere.

Important Questions on Electric Charges and Fields

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A charged wire $A B$ of length $l$ is placed inside a sphere, as shown in figure. Linear charge density of wire is $λ = k x$ , where $x$ is the distance measured along the wire from end $A$ and $K$ is constant. Calculate the flux through the sphere.