# NCERT Exemplar Solution for CBSE Class 12 Physics, Chapter 1: Electric Charges and Fields (Part I)

NCERT Exemplar solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields are available here. This chapter has a large number of questions due to which we have provided the solutions in several parts. This is Part I, here you will find solutions for **MCQs I** (or multiple choice questions with single correct answer). Solutions for rest of the questions are available in other parts. These questions are very important for CBSE 12th Physics board exam & competitive examinations like JEE Main, NEET etc.

*NCERT Exemplar Solution for 12th Physics, Chapter 1: Electric Charges and Fields (MCQs I) are given below:*

**Question 1.1:** In figure two positive charges *q*_{2} and *q*_{3} fixed along the *y*-axis, exert a net electric force in the + *x*-direction on a charge *q*_{1}, fixed along the *x*-axis. If a positive charge Q is added at (*x*, 0), the force on *q*_{1},

**(a)***shall increase along the positive x-axis*

**(b)*** shall decrease along the positive x-axis*

**( c)** shall point along the negative x-axis

**( d)** shall increase but the direction changes because of the intersection of Q with

*q*

_{2 }and

*q*

_{3}

**CBSE Class 12 Syllabus 2017-2018: All Subjects**

**Solution 1.1: ( a)**

Net force on charge *q*_{1}, by other charges *q*_{2} and *q*_{3} is along the + x-direction, so nature of force between *q*_{1} & *q*_{2} and *q*_{1} & *q*_{3} is attractive. This is possible when charge *q*_{1} is negative.

Now, if a positive charge *Q *is placed at (*x*, 0), then, the force on *q*_{1} shall increase. The direction will be along positive *x* axis.

**NCERT Exemplar Problems' PDF: Class 12 Physics – Chapter 1 Electric Charges and Fields**

**NCERT Solutions for Class 12 Physics ‒ Chapter 1: Electric Charges and Fields (Part I)**

**Question 1.2:** A point positive charge is brought near an isolated conducting sphere (figure). The electric field is best given by

**( a)** Fig (i)

**( b)** Fig (ii)

**( c)** Fig (iii)

**( d)** Fig (iv)

**Solution 1.2: ( a)**

Electric field originates from positive charge. It is perpendicular to the surface of the conductor. Also, inside the conducting sphere, it will be zero.

All these conditions are satisfied in Fig (i).

**Question1.3:** The Electric flux through the surface

**( a)** in Fig.1.3 (iv) is the largest.

**( b)** in Fig. 1.3 (iii) is the least.

**( c)** in Fig. 1.3 (ii) is same as Fig. 1.3 (iii) but is smaller than Fig. 1.3 (iv)

**( d)** is the same for all the figures.

**Solution 1.3: ( d)**

The flux of electric field through any closed surface S is 1/ε_{0} times the total charge enclosed by S (Irrespective of shape and size of the surface).

All the figures given in the questions have the same charge i.e., +*q*. So, the electric flux through the surface is the same for all the figures.

**Question1.4:** Five charges *q*_{1}, *q*_{2}, *q*_{3}, *q*_{4}, and *q*_{5} are fixed at their positions as shown in Fig. 1.4. S is a Gaussian surface.

The Gauss’s law is given by

Which of the following statements is correct?

**( a)**

**E**on the LHS of the above equation will have a contribution from

*q*

_{1},

*q*

_{5 }and

*q*

_{3}while

*q*on the RHS will have a contribution from

*q*

_{2}and

*q*

_{4}only.

**( b)**

**E**on the LHS of the above equation will have a contribution from all charges while

**q**on the RHS will have a contribution from

*q*

_{2}and

*q*

_{4}only.

**( c)**

**E**on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from

*q*

_{1},

*q*

_{3}and

*q*

_{5}only.

**( d)** Both

**E**on the LHS and

*q*on the RHS will have contributions from

*q*

_{2}and

*q*

_{4}only.

**Solution 1.4: ( b)**

According to Gauss' theorem, the term* q* on the right side of the equation includes the sum of all charges enclosed by the surface (irrespective of the position of the charges inside the surface).

If the surface is so chosen that there are some charges inside and some outside, the electric field on the left side of equation will be due to all charges (both inside and outside)

Here, **E** on LHS of the above equation will have a contribution from all charges while *q *on the RHS will have a contribution from charges *q*_{2} and *q*_{4} only.

**Question1.5:** Figure 1.5 shows electric field lines in which an electric dipole** p** is placed as shown. Which of the following statements is correct?

**( a)** The dipole will not experience any force.

**( b)** The dipole will experience a force towards right.

**( c)** The dipole will experience a force towards left.

**( d)** The dipole will experience a force upwards.

**Solution 1.5: ( b)**

We know electric field emerges radially outward from positive point charge.

In the figure given above, space between field lines is increasing (or density of electric field line is decreasing). In other words, the electric force is decreasing while moving from left to right.

Thus, the force on charge ‒*q* is greater than the force on charge + *q* in turn dipole will experience a force towards left direction.

**Question** **1.6:** A point charge +*q*, is placed at a distance *d* from an isolated conducting plane. The field at a point *P* on the other side of the plane is

**( a)** directed perpendicular to the plane and away from the plane.

**( b)** directed perpendicular to the plane but towards the plane.

**( c)** directed radially away from the point charge.

**( d)** directed radially towards the point charge.

**Solution 1.6: ( a)**

When you place a positive charge near a conducting plane, then electric field lines from positive charges will enter into the conducting plane (from the side where positive charge is kept) and emerge from opposite side of the plane.

In both cases, the direction of electric field lines will always be perpendicular to the surface of the plane.

**NCERT Solution for Class 12 Physics: Chapter 1: Electric Charges and Field (PDF Format)**