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ELECTRICAL CHARGES and FIELDS– NUMERICAL PROBLEMS
1. How many electrons must be removed from a body to make its charge 5mC? Ans.: From quantisation of charge: q = ne n =
(^316) 19 q (^5 10) 3.125 10 e 1.6 10
2. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7^ C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene? Ans: (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged. Amount of charge on the polythene piece, q = − 3 × 10−7^ C Amount of charge on an electron, − e = −1.6 × 10−19^ C Number of electrons transferred from wool to polythene = n (should be positive) WKT, From quantisation of charge: q = ne but here we take q = − ne n =
(^712) 19 q (^3 10) 1.875 10 e 1.6 10
The number of electrons transferred from wool to polythene is 1.875 × 10^12. (b) Yes. There is a transfer of mass taking place. This is because an electron has mass, me = 9.1 × 10−3^ kg Total mass transferred to polythene from wool, m = me × n = 9.1 × 10−31^ × 1.875 × 10^12 = 17.06 × 10−1^9 kg Hence, a negligible amount of mass is transferred from wool to polythene.
3. What is the force between two small charged spheres having charges of 2 × 10−^7 C and 3 × 10−^7 C placed 30cm apart in air? Ans: Given charge on the first sphere, q 1 = 2 × 10−^7 C Charge on the second sphere, q 2 = 3 × 10−^7 C Distance between the spheres, r = 30 cm = 0.3 m Electrostatic force between the spheres is given by the relation,
0 1 22
F =^1 q q 4 πε r 9 7 7 5 3 2 F = 9×10 2 ×10^ 3 ×10^9 6×10 6×10 N 0.3 0 09_._
^ ^ ^ Hence, force between the two small charged spheres is 6 × 10−^3 N. The charges are of same nature. Hence, force between them will be repulsive.
4. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Ans: (a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q 1 = 0.4 μC = 0.4 × 10−^6 C Charge on the second sphere, q 2 = − 0.8 μC = − 0.8 × 10−^6 C
Electrostatic force between the spheres is , 0 1 22
1 q q F = 4 πε r
2 1 2 9 6 6 3 0
r = 1 q q 9×10 0.4×10^ ×0.8×10 14 4 × 10 0 0144 4 πε F 0 2_..^._
^ (^) r 0 0144_._ 0 12m_._ The distance between the two spheres is 0.12 m = 12cm. (b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
6. Two point charges q A = 5 μC and q B = − 5 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 2 × 10−9^ C is placed at this point, what is the force experienced by the test charge?
Ans: (a) The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, AB = 20 cm , ∴ AO = OB = 10 cm
a) Electric field at O due to charge qA : A^0 A^2
E 1 q
4 r
Magnitude of electric field at O due to each charge is 9 6 5 1 EA EB 9×10 5×100 1_._ 2 45 × 10 NC
Net electric field at O is
E EA E B90 × 10 N C^5 ^1 = 9 × 10 N C^6 ^1 along OB
b) Magnitude of force experienced by the test charge: F q E F 2 × 10 ^9 × 90 × 10^5 180 × 10 ^4 N = 1.8 × 10 ^2 N along OA
Therefore, the electric field at mid-point O is 9 × 10^6 N C−1^ along OB.
7. Two insulated charged spheres of charges 6.5 × 10 −^7 C each are separated by a distance of 0.5m. (i) Calculate the electrostatic force between them. (ii) Also calculate the force (a) When the charges are doubled and the distance of separation is halved. (b) When the charges are placed in a dielectric medium water (εr = 80).
Ans.: (i) Electrostatic force: 0 1 22
F 1 q q
4 r
9 7 7 2 5 2
F 9 × 10 6.5 × 10^ ×6.5 × 10
F 1521 × 10 = 1.52 × 10 N
^
(ii) (a) Now q 1 = q 2 = 2 × 6.5 × 10 −^7 = 13× 10 −^7 C , r = 0.5/2 = 0.25m
New electrostatic force: 0 1 22
F 1 q q
4 r
9 7 7 2 5
F 9 × 10 13 × 10^ ×13 × 10
F 24336 × 10 = 0.243 N
^
(b) When the charges are placed in a dielectric medium water
New electrostatic force: 1 2 2 0 r 2 r
F 1 q q F^ 1.52 × 10
4 r 80
0 019× 10. 2 N 1 9× 10. ^4 N
11. Two fixed point charges +4μC and +1μC are separated by 30cm in air. Find the position between them at which resultant electric field is zero. Ans.: Given, charge: q 1 = 4 μC = 4×10−^6 C Charge: q 2 = 1 μC = 10−^6 C Distance between the charges is r = 30cm = 0.3m Electric field produced by a point charge at a distance x is
0 2
E 1 q
4 x
Now, the net electric field at a point (P) between the two charges is zero. The magnitudes of the fields produced by q 1 and q 2 are equal in magnitude and opposite in direction. i.e., E 1 = E 2
(^12 ) 0 0
1 q 1 q
4 x 4 r x
6 6 2 2
4 ×10 10
x 0 3. x
4 0 3. x (^) ^2 x^2 Taking square root on both sides, 2(0.3 – x) = x Solving, 0.6 – 2x = x OR 0.6 – 2x = – x i.e., 0.6 = 3x x = 0.2 m OR 0.6 = x But x = 0.6 m lies outside the line joining the charges. Thus, at a distance 0.2m from +4μC between the line joining the charges the net electric field is zero.
13. In an electric dipole, the charges are +3 μC and – 3 μC separated by a distance 6mm. Calculate the magnitude of electric field due to the dipole at a point on the (i) axis at a distance 4mm from the midpoint and (ii) equatorial line at a distance 3mm from the midpoint. Ans: Given, magnitude of each charge: q= 3×10–^6 C, Separation or distance between charges : 2a = 6mm = 6×10–^3 m, and a = 3mm = 3×10–^3 m The dipole moment: p = 2aq = 6 ×10–^3 × 3 × 10–^6 = 18 × 10–^9 C m (i) Here distance: r = 4mm = 4 ×10–^3 m Magnitude of the electric field produced by the dipole at a point on the axis
at a large distance r is^ ax^0 2 2 ^2
E 1 2 p r
4 r a
9 9 3 ax (^3 2 3 )
E 9 × 10 ×^ 2 × 18 × 10^ × 4 × 10
(4 ×10 ) (3 ×10 )
Eax = 2.64×10^10 N/C or V/m
(ii) Magnitude of the electric field produced by the dipole at a point on
equatorial line at a large distance r is
eq (^2 2 ) 0
E 1 p
4 r a
9 9 eq (^3 2 3 2 )
E 9 × 10 ×^ 18 × 10
(4 ×10 ) (3 ×10 )
Eeq = 6.36×10^6 N/C or V/m
14. A positive charge 10 μC is placed 1m away from a similar negative charge in vacuum. Calculate the electric field at a point 1m away from each charge. Ans.: Given magnitude of each charge: q = 10 μC = 10 ×10–^6 C Distance between charges: d = 1 m Distance: AP = BP = d = 1m The magnitude of the electric field at P due to the charge ‘+q ’ is (^1 )
E =^1 q
4 πε d
9 6 4 (^1 )
E 9 × 10 × 10 × 10 9 × 10 N/C
The magnitude of the electric field at P due to the charge ‘– q’ is
(^2 ) E =^1 q 4 πε (^) d 9 6 4 (^2 )
E 9 × 10 × 10 × 10 9 × 10 N/C
Resultant electric field at P, E^ ^ E 1 2 ^ E 2 2 ^ 2E E 1 2 ^ cos
E 9 × 10^4 2 9 × 10^4 2 2 9 × 10^4 9 × 10 4 cos120o 9 × 10^4 N/C
The resultant electric field makes an angle of 60o^ with E 1 as well as E 2.
