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Class 12 Physics Chapter 1 Electric Charges and Fields Ques & Ans

CBSE Class 12 Physics Chapter 1 Electric Charges and Fields Questions Answer for MCQs

Subject matter experts of Result4u have developed chapter wise practice questions of Class 12th as per the relevancy of the chapter in the CBSE board exam 2023. This practice set contains 8 questions with detailed explanations on “Chapter: Electric Charges and Fields” of Physics MCQs.

  • Ques1.Figure shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is correct?
    electric field lines

Explanation: The space between the electric field lines is increasing , here from left t right and its characteristics states that , strength of electric field decreases with the increase in the space between electric field lines .as a result force on charges also decreases from left to right.
Thus , the force on charge – is greater than the force on charge +q in turn dipole will experience a force towards left.

Ques 2: The Electric field at a point is

Explanation: The electric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed at that pint . thus electric field due to the charge Q will be continuous , if there is n charge at that pint .it will be discontinuous if there is acharge at that pint.

Ques3.If there were only one type of charge in the universe, then
(i) ∫ E.dS≠ 0 on any surface.
(ii) ∫ E.dS= 0 if the charge is outside the surface.
(iii)∫ E.dS could not be defined.

(iv)E.dS=qε0,ifchargesofmagnitudeqwereinsidethesurface.(iv) ∫E.dS =qε0,if charges of magnitude q were inside the surface.
Explanation:
Gauss law states that ) E.dS=qε0,∫E.dS =qε0, where q is the charge enclosed by the surface. If the charge is outside the surface, then charge enclosed by the surface is q=0 and thus ) ∫ E.dS = 0.Here electric flux does not depends on the type or nature of charge.

Ques4.Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region

Explanation: When there are various types f charges in a region , but the total charge is zero, the region can be supposed to contain a number of electric dipoles.
Therefore , at points outside the region, the dominant electric field is ∝ 1/r3,for large r, where r is the distance from a origin in this region
Further , as electric field is conservative , work done to move a charged particle along a clsed path away from the region is zero.
Ques5.Refer to the arrangement of charges in Fig. and a Gaussian surface of radius R with Q at the centre. Then
surface of spher

Explanation: Gauss law states that the total electric flux of an enclosed surface is given by)

E.dS =qε0,∫E.dS =qε0,

where q is the charge enclosed by the surface
Thus from figure,
Total charge inside the surface is Q -2Q =-Q
Total flux through the surface of the sphere

=Qε0=−Qε0

Now considering charge 5Q . Charge 5Q lies outside the surface, thus it makes no contribution to electric flux due to given surface.

Ques 6.A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring (Fig.). Then
centre of the ring

Explanation:
Gauss law states that E.dS =qε0,∫E.dS =qε0, where q is the charge enclosed by the surface.
If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.
If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.
If q < 0, it will perform SHM for small displacement along the axis.

Ques 7.A point positive charge is brought near an isolated conducting sphere (Fig). The electric field is best given by
isolated conducting sphere

Explanation: When a positive point charge is brought near an isolated conducting sphere without touching the sphere , then the free electrons in the sphere are attracted towards the positive charge. This leaves an excess of positive charge on the rear surface of sphere.
Both kinds of charges are bund ij the metal sphere and cannot sescape. They , therefore, reside n the surface.
Thus, the left surface of sphere has an excess of negative charge and the right surface of sphere has an excess of positive charge .
As electric filed lines start from posituive charge and ends at negative charge .
Als electric field linne emerges from a positive charge , in case f single charge and ends at infinity.
Here all these conditions are fulfilled in fig (i)

Ques 8.Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in Fig. S is a Gaussian surface. The Gauss’s law is given by SEds=q0∮SE⋅ds=q∈0

Which of the following statements is correct?

Explanation: According to Gauss law , the term q on the right side of the equation includes the sum of all charges enclsed by the surface.
The charges may be located anywhere inside the surface , if the surface is so chosen that there are sme charges inside and some utside , the electric field n the left side of equation is due t all the charges , both inside and outside S.
So,E on LHS of the above equation will have a contribution from all charges while q on the RHS will have a contributin from q2 and q4 nly.

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