Drive the expression for electric field at a point on the equato

Subject

Physics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1.

Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give a reason for your answer.

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2.

A long straight current carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will the be an induced emf in the loop? Justify

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3.

At a place, the horizontal component of earth's magnetic field is B and angle of dip are 60o. What is the value of a horizontal component of the earth's magnetic field at the equator?


4.

How is the speed of em-waves in vacuum determined by the electric and magnetic field?

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5.

How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. 

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6.

(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage?

(ii) Without making any other change, find the value of the additional capacitor C, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.

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7.

A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of Ro. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire.

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8.

(i) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2μF capacitance.

(ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?

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9.

Drive the expression for electric field at a point on the equatorial line of an electric dipole.


The magnitudes of the electric field due to the two charges +q and -q are given by,
straight E subscript plus straight q end subscript space equals space fraction numerator straight q over denominator 4 space πε subscript 0 end fraction fraction numerator 1 over denominator straight r squared space plus straight a squared end fraction space..... space left parenthesis straight i right parenthesis
straight E subscript negative straight q end subscript space equals space fraction numerator straight q over denominator 4 πε subscript 0 end fraction fraction numerator 1 over denominator straight r squared plus straight a squared end fraction space space....... space left parenthesis ii right parenthesis

The directions of E+q and E-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
Therefore, Total electric field.
E = - (E+q + E-q) cosθ p [Negative sign shows that field is opposite to p]

straight E space equals space minus fraction numerator 2 qa over denominator 4 πε subscript 0 space left parenthesis straight r squared space plus straight a squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction space space space... left parenthesis iiii right parenthesis
At large distances (r>>a),this reduces to
straight E space equals negative fraction numerator 2 qa over denominator 4 πε subscript 0 straight r cubed end fraction straight p with hat on top space space.... left parenthesis iv right parenthesis
because space straight p with rightwards arrow on top space equals space straight q space straight x space 2 straight a straight p with hat on top
therefore space straight E space equals space fraction numerator negative straight p with rightwards arrow on top over denominator 4 πε subscript 0 straight r cubed end fraction space left parenthesis straight r greater than greater than straight a right parenthesis

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10.

Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field.

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