Kendriya Vidyalaya Kv No 1 Shahibaug, Ahemadabad
KENDRIYA VIDYALAYA KV NO 1 SHAHIBAUG, AHEMADABAD
WORKSHOP ON 02/07/2016
QUESTION BANK FOR ELECTROSTATICS
1. / Name the physical quantity whose S.I. unit is J/C. Is it a scalar or a vector quantity?Ans: (i)electric potential (P.d), (ii) scalar quantity / 1
2. / How does a torque affect the dipole in an electric field?
Ans: Torque to align the dipole along the electric field / 1
3. / What is the work done in moving a charge 10 nC Between two point on equipotential surface?
Ans: No work is done to move charge on equipotential surface. / 1
4. / Calculate the capacity of sphere of radius of 10km.
Ans: C = 4ε0r = 1.1μF / 1
5. / Is the force acting between two point electric charges q1 and q2, kept at some distance apart in air, attractive or repulsive, when (i) q1q2 > 0 (ii) q1q2 <0 ?
Ans: (i) The force is repulsive. When q1q2 > 0, (ii) The force is attractive. When q1q2 < 0, / 1
6. / You are given three capacitors of value 2µF, 3µF, 6µF. How will you connect them to a resultant capacity of 4µF?
Ans: 3µF, 6µF are connected in series and 2µF is connected in parallel / 2
7. / An uncharged insulated conductor A is brought near a charge insulated conductor B what happens to charge and potential of B?
Ans: (i) charge on B conductor remain same (ii) potential of B get lowered because induces charge of opposite side of conductor A / 2
8. / Find the ratio of potential difference that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitance in the ratio 1:3 so that the energy stored in the two cases is same.
Ans: Given
Substituting CS-And Cp= C1+C2
/ 2
9. / (i) Can two equipotential surfaces intersect each other? Give reasons.
(ii) Two charges −q and +q are located at points A (0, 0, −a) and B (0, 0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (−3, 0, 0)?
Solution:
(i) Two equipotential surfaces cannot intersect each other because when they will intersect, the electric field will have two directions, which is impossible.
(ii) Charge P moves on the perpendicular bisector of the line joining +q and −q. Hence, this perpendicular bisector is equidistant from both the charges. Thus, the potential will be same everywhere on this line. Therefore, work done will be zero. / 2
10. / A spherical Gaussian surface encloses a charge of 8.85 × 10-10C.
(i) Calculate the electric flux passing through the surface.
(ii) How would the flux change if the radius of the Gaussian surface is doubled and why?
Solution:
= 100 T m2
- Total flux enclosed 100 NC-1 m2
- The flux would not change if the radius of Gaussian surface is double because enclosed charge remains the same.
11. / Three charges –q, +Q and –q are placed at equal distance on a straight line. If the potential energy of the system of three charges is zero, find the ratio of Q:q
Ans: As total potential energy is zero
/ 3
12. / Four point charges are placed at the four corners of a square in the two ways (i) and (ii) as shown below. Will the (i) electric field (ii) Electric potential, at the centre of the square, be the same or different in the two configurations and why?
ANS
(I) Electric field is a vector quantity .in the first case electric field at the center due to charges at A and due to C adds up and also due to charges at B and D are added up. There exists electric field at the centre. Where as in the second case field due to A and C are equal and opposite and also due to B and D are also equal and opposite, so the resultant field is zero.
(ii) Potential is a scalar quantity and is positive due to positive charge and negative due to negative charge. Hence resultant potential at the centre is zero in both the cases / 3
13. / An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.
Solution:
(i) Consider an electric dipole consisting of two equal and opposite point charges, −q at A and + q at B, separated by a small distance 2a.
AB = 2a, having dipole moment
Let this dipole be held in a uniform external electric field at an angle with the direction of .
Force on charge −q at A = − q , in a direction opposite to
Force on charge +q at B = +q , along the direction of
Net force on the dipole = qE − qE = 0
(ii) Work done on dipole, W = U = P E (cos 1 − cos 2
W = P E (cos0° − cos180°)
W = 2pE / 3
14 / A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor
Solution:
(i) Q = CV
Therefore, the electric field between the parallel plates depends only on the charge and the plate area. It does not depend on the distance between the plates.
Since the charge as well as the area of the plates does not change, the electric field between the plates also does not change.
(ii)Let the initial capacitance be C and the final capacitance be C'.
Accordingly,
Hence, the capacitance of the capacitor gets halved when the distance between the plates is doubled.
Energy of a capacitor, U
Since Q remains the same but the capacitance decreases,
The energy stored in the capacitor gets doubled when the distance between the plates is doubled. / 3
15. / Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point
(i) outside and (ii) inside the shell.
Plot a graph showing variation of electric field as a function of r > R and r < R.
(r being the distance from the centre of the shell)
OR
Electric Field Due To A Uniformly Charged Thin Spherical Shell:
(i) When point P lies outside the spherical shell:
Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let be the electric field at point P. Then, the electric flux through area element is given by,
Since is also along normal to the surface,
dΦ = E ds
∴ Total electric flux through the Gaussian surface is given by,
Now,
Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,
From equations (i) and (ii), we obtain
(ii) When point P lies inside the spherical shell:
In such a case, the Gaussian surface encloses no charge.
According to Gauss law,
E × 4πr2 = 0
i.e., = E = 0 (r < R)
Graph showing the variation of electric field as a function of r:
/ 3
16 / Derive an expression for dipole field intensity at any point on
(i)axial line of dipole
(ii) equatorial line of dipole
Ans:
Electric field intensity at a point on axial line: -
Consider a dipole of charge +q & q separated by a distance 2l so dipole moment p = 2ql
We have to find field intensity at point P, at distance r from centre of the dipole. So electric field due to +q charge is {along OP}
Intensity at P due to –q is{along PO}
So resultant intensity at P is E = E1 + E2 =
So
So
Therefore Since p = q×2l , is electric dipole moment.
Special case: If dipole is very small, then r > l so r2 >l2, therefore neglecting r2.
Therefore In vector form .
Electric Field at a point on equatorial line of a dipole:- Consider a dipole with charges + q & q separated by a distance of 2 l. So dipole moment p = 2 q l - - - -(1)We have to find intensity of electric field at point A on the equatorial line of the dipole at distance ‘r’ from the centre of the dipole. So
Electric field intensity at A due to +q charge is along BA
------(2) Electric field at A due to –q charge is
- - - - (3) {Along AC} Since
. Since the vertical components are equal but opposite so the cancels out each other but the horizontal component are along same direction so resultant field intensity at A due to dipole is Therefore
Therefore E =
Where P= 2ql is dipole moment. Its direction is parallel to the dipole axis and from +ve to –ve.
Special case:- For a small dipole 2l is very small so l < r hence l2 < r2. Now we can neglect l2.
E = E = / 5
16 / State gauss’s law in electrostatics. Use this law to drive an expression for the electric field due to an infinitely long straight wire of linear charge density .
Ans
The surface integral of the normal component of the electric field over any closed hypothetical surface is equal to 1/0 total charge enclosed by the surface”.
OR
“The net outward electric flux through any closed hypothetical surface is equal to 1/0 total charge enclosed by the surface”.
Electric Field Intensity due to two uniformly Charged infinite long Charged Conductor:-Consider a thin infinite long positively charged straight long conductor of linear charge density
= q / l - - - -(1).
We have to find electric field intensity at point P, which is at distance ‘r’ from the plate.
With conductor as an axis we will draw an imaginary cylinder of radius ‘r’ & length ‘l’ it is called Gaussian surface & point ‘P’ lies on it. Electric field at each point on the conductor is same & directed perpendicularly outward.
From Gauss’s theorem the outward flux through the Gaussian surface is
(i)Flux through the plane surfaces A & B is minimum because area vector is perpendicular to the field i.e. = 90o
hence minimum lines pass through it ------( 1 )
Due to same reason flux through plane surface B is also zero.
( ii )Flux through the curved surface is maximum because area vector is parallel to the field i.e. = 0 hence maximum lines pass through it.
Hence net outward flux through the Gaussian surface is = 1 + 2 + 3
From eq 1 & 2,
From Gauss’s theorem Comparing eq3 & 4
Hence electric field at a point near a straight long conductor is inversely proportional to distance of the point. / 5