## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Electric Charges and Fields

#### Multiple Choice Questions

11.

Two parallel large thin metal sheets have equal surface charge densities ( σ = 26.4 × 10-12 C/m2 ) of opposite signs. The electric field between these sheets is

• 1.5 N/C

• 1.5 × 10-10 N/C

• 3 N/C

• 3 × 10-10 N/C

C.

3 N/C

Electric field  between the sheets is

E =  $\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_{0}}$

=

Since given:- ε0 = 8.85 × 10-12 C2 / (Nm2)

E  ≈ 3 N/C

12.

The figure shows three points A, B and C in a region of a uniform electric field $\stackrel{\to }{\mathrm{E}}$. The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good?

• VA = VB = VC

• VA = VB > VC

• VA = VB < VC

• VA > VB = VC

B.

VA = VB > VC

Electric lines of force flow from higher potential to lower potential so,

VA = VB > VC

13.

If an insulated non-conducting sphere of radius R has charge density ρ. The electric field at a distance r from the centre of sphere (r > R) will be

C.

Gauss law allows us to determine the electric field at a point when there is sufficient symmetry.

Charge which is enclosed in the surface is given by

14.

Number of extra electrons in a body of charge -80$\mathrm{\mu }$C, is

• 80 × 1015

• 80 × 10−15

• 5 × 1014

• 1.28 × 10−17

C.

5 × 1014

Charge on an electron = 1.6 × 10−19 C

Number of extra electrons in an object having charge −80$\mathrm{\mu }$C

=

= 5 × 1014

15.

Two spheres of radii R1 and R2 respectively are charged and joined by a wire. The ratio of the electric field of spheres is :

• $\frac{{\mathrm{R}}_{2}^{2}}{{\mathrm{R}}_{1}^{2}}$

• $\frac{{\mathrm{R}}_{1}^{2}}{{\mathrm{R}}_{2}^{2}}$

• $\frac{{\mathrm{R}}_{2}}{{\mathrm{R}}_{1}}$

• $\frac{{\mathrm{R}}_{1}}{{\mathrm{R}}_{2}}$

C.

$\frac{{\mathrm{R}}_{2}}{{\mathrm{R}}_{1}}$

The spheres are joined by the wire, so their potential will become equal.

V1 = V2

= R2/R1

16.

A conducting sphere of radius R = 20 cm is given by a charge Q = 16μC. What is E at centre?

• 3.6×106 N/C

• 1.8×10N/C

• Zero

• 0.9×106 N/C

C.

Zero

The value of E at the centre of the sphere is zero. Since the centre of the sphere is placed within the Gaussian surface. The use of Gauss law to examine the electric field of a charged sphere shows the electric field enviornment outside the sphere is identical to that of point charge.

Therefore the potential is the same as that of a point charge

hence E=0

17.

A charge q is located at the centre of a cube. The electric flux through any face is

• $\frac{\mathrm{\pi q}}{6\left(4{\mathrm{\pi \epsilon }}_{0}\right)}$

• $\frac{\mathrm{q}}{6\left(4{\mathrm{\pi \epsilon }}_{0}\right)}$

• $\frac{2\mathrm{\pi q}}{6\left(4{\mathrm{\pi \epsilon }}_{0}\right)}$

A.

According to Gauss's theorem, electric flux through the cube (closed surface),

Since, cube has six surfaces and all the faces are symmetrical, therefore electric flux through any face

18.

Two charged spheres separated by a distance d exert some force (F) on each other. If they are immersed in a liquid of dielectric constant 4, then what is the force exerted, if all other conditions are same ?

• 2F

• 4F

• $\frac{\mathrm{F}}{2}$

• $\frac{\mathrm{F}}{4}$

D.

$\frac{\mathrm{F}}{4}$

19.

The spatial distribution of the electric field due to two charges (A, B) is shown in figure. Which one of the following statements is correct?

• A is +ve and +B is -ve and

• A is $-$ve and B is +ve;

• both are +ve but A > B

• both are $-$ve but A > B

A.

A is +ve and +B is -ve and

The lines proceed from the positive and lines will reach B, if B is negative.

The number of lines/unit area i.e the intensity of lines is great the charge.

A is positive, B is negative and

20.

In a discharge tube ionization of enclosed gas is produced due to collisions between

• positive ions and neutral atoms/molecules

• negative electrons and neutral atoms/molecules

• photons and neutral atoms/molecules

• neutral gas atoms/molecules

B.

negative electrons and neutral atoms/molecules

In a discharge tube, after being accelerated though a high potential difference the ions in the gas strike the cathode with huge kinetic energy. This collision liberates electrons from the cathode. These free electrons can further liberate ions from gas molecules through collisions. The positive ions are attracted towards the cathode and negatively electrons move towards anode. Thus, ionization of gas results.

840 Views