INDIAN SCHOOL, ALWADI ALKABIR
DEPARTMENT OF SCIENCE 2016 -17
CLASS – XII PHYSICS DATE- 2/4/2016
WS. NO. 1 UNIT1-ELECTROSTATICS
Electric field and electric potential
[1] In a certain 0.5 3m of space, potential is found to be 7V throughout.What is the E.F?[zero]
[2]Is electrostatic potential necessarily zero at point where electric field ‘E’ is zero? Why[no]
[3] What should be the work done if a point charge + q is taken from a point A to the point B on the circumference drawn with another point + q at the centre ?
[4]What is the unit of electric flux. A charge ‘q’ is enclosed by a spherical surface of radius ‘R’ .If the radius is reduced to half ,how would the electric flux through the surface changes? Nm2C-1or Vm
[5]In the fiq. q1, q2 and q3 are the 3 charges. A Gaussian surface is drawn around q1 andq2. [i]What is the net charge which determines the electric flux through the Gaussian surface? [ii] Which all charges produce the electric field at the point ‘P’?
[Hint: Use Gauss law]
[6]Two point charges of 2 μC& 8 μC are placed 12cm apart. Find the position of the point where the electric field intensity will be zero[4cm from 2 μC]
[Hint:E1 = E2 at the given point]
[7]A hollow metal sphere of radius 5cm is charged such that the potential on its surface is 20V.What is the {i}potential at its centre of the sphere.{ii} electric field intensity at the centre[20V and zero]
[8]S1 and S2 are two hollow concentric spheres enclosing charges 2Q and 4Q respectively as shown in the figure.
What is the ratio of electric flux through S1 andS2?
[Hint:Use Gauss law]
[9]Charges +q and –q are placed at points A and B respectively ,which are a distance 2L apart . C is the mid point between points A and B .Find the work done in moving a charge +Q along the semicircle CRD
Hint: p.d = work x charge
.
[10Two charges 20 μc and -20 μc are placed 20 cm apart at points A and B respectively, compute the electric field at C, 20 cm away from both A and B.
Hint:Use superposition principle
[11]Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of 1cm.SurfaceA is given a positive potential of 10v and the outer surface of B is earthed.[i] What are the magnitude and direction of the electric field between the plates[ii] What is the work done in moving a charge of 20 µc from a point on the plate A to another point on the plate B?
E = 10/.01= 1000N/c[ii] w = q v = 20 x 10-5 J
[12] A capacitor each with plate area A and separation d is charged to potential
difference V .The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant ‘K ‘is now placed between the plates. What change if any take place [i] charge on the plates[ii] capacitance of the capacitor [iii] p.d between the plates[iv] electric field intensity between the plates[v] energy stored in the capacitor [i] q remains the same[ii] capacitance increases as Cd= KCo[iii] potential difference decreases [iv] electric field intensity decreases [v] energy decreases
[13]A parallel plate capacitor ’C’ remains connected across a supply voltage of ‘V’ volts.Now a dielectric slab of constant ‘K’ fills the space between the plates. What will be the new [i] p.d [ii] capacitance [iii] charge stored[iv] electric field [v] energy stored.
[i]same[ii] increase K times[iii increases k times[iv] E1 =V/d ,remains the same[v] U1 = ½ C1 x V2 = ½ KC V2 ,increases K times
Prepared by Mr. William
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