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Class 10 Class 12

Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10^–7 C? The radii of A and B are negligible compared to the distance of separation.

C harge on sphere 1, q_{1} = 6.5 \times 10^{- 7} C C harge on sphere 2, q_{2} = 6.5 \times 10^{- 7} C D istance between the centre of spheres, r = 50 cm = 0.50 m k = \frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2} Using Coulmob' s law for Electrostatic force

F = k \frac{q_{1} q_{2}}{r^{2}}

= \frac{9 \times 10^{9} \times 6.5 \times 10^{- 7} \times 6.5 \times 10^{- 7}}{(0.50)^{2}} N

F = 1.5 \times 10^{- 2} N

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A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10^–7C.
a) Estimate the number of electrons transferred (from which to which?)
b) Is there a transfer of mass from wool to polythene?

Given, Charge - q = - 3.2 \times 10^{7} C and charge on electron - e = - 1.6 \times 10^{- 19} C ∴ number of electrons transferred q = ne n = \frac{q}{e} = \frac{- 3.2 \times 10^{- 7}}{- 1.6 \times 10^{- 19}} = 2 \times 10^{12}

b)
Electrons are transferred from wool to polythene during rubbing as polythene has negative charge.

Mass of an electron, m =

9.1 \times 10^{- 31} kg

No. of charge, n =

2 \times 10^{12}

Therefore, amount of mass transferred= = 2 x 10¹² x 9.1 x 10^–31 kg = 18.2 x 10^–19 kg.

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Suppose the spheres A and B in question 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Charge on each of the sphere A and B = q = 6.5 x 10^–7 C

When a similar but uncharged sphere C is placed in contact with sphere A, each sphere shares a charge q/2, equally.
Now, if the sphere C is placed in contact with sphere 8, the charge is equally redistributed, so that
Charge on sphere B or C =

1 half left parenthesis straight q plus straight q divided by 2 right parenthesis space equals space 3 straight q divided by 4

Thus, the force of repulsion between A and B is

straight F space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction. fraction numerator begin display style fraction numerator 3 straight q over denominator 4 end fraction end style. straight q divided by 2 over denominator left parenthesis straight r divided by 2 right parenthesis squared end fraction space space space space equals space 3 over 8. space fraction numerator 1 over denominator 4 πε subscript 0 end fraction. straight q squared over straight r squared space space space space space equals space 3 over 8 cross times 1.5 space cross times space 10 to the power of negative 2 end exponent straight N space space space space equals space 0.5625 space cross times space 10 to the power of negative 2 end exponent straight N space space space space equals 5.7 space cross times space 10 to the power of negative 3 end exponent straight N.

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Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Since particles (1) and (2) are deflecting towards the positively charged plate both of these particles are negatively charged. Particle (3) is positively charged because it is getting deflected towards negatively charged plate.

Deflection experienced by a particle is directly proportional to charge/mass ratio.
Therefore, since particle 3 experiences maximum deflection it has the maximum charge to mass ratio.

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Electric Charges and Fields

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?