What kind of charges are produced on each when (i) a glass rod is rubbed with silk and (ii) an ebonite rod is rubbed with wool?

(i) Positive charge will be produced on glass rod and negative charge will be produced on silk. The electrons are less tightly bound in glass rod as compared to silk.

(ii) Negative charge will be produced on ebonite rod and positive charge will be produced on wool.

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Two charged particles having charge 2.0 x 10-8 C each are joined by an insulating string of length 1 m and the system is kept on a smooth horizontal table. Find the tension in the string.


Here given the particles have an equal charge joined by a string of length 1 m.

                   q1 = q2 = 2 × 10-8Cr = 1 m

Tension in the string is the force of repulsion (F) between the two charges.

According to Coulomb's law,
                 F = q1 q24π 0 r2   =9 × 109 (2 × 10-8) (2 × 10-8)12
         
                 F = 3.6 × 10-6N


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A free pith ball of mass 8 g carries a positive charge of 5 x 10–8 C. What must be the nature and magnitude of charge that should be given to a second pith ball fixed 5 cm vertically below the former pith ball so that the upper pith ball is stationary?

Let m be the mass of the upper ball. Let q1 represent the charge on the upper ball. For equilibrium of upper ball, it must experience an upward electric force Fe. This is possible if the lower ball has positive charge, say, q2.
For equilibrium,   fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 space straight q subscript 2 over denominator straight r squared end fraction space equals space mg
Substituting values,
     fraction numerator 9 cross times 10 to the power of 9 cross times 5 cross times 10 to the power of negative 8 end exponent cross times straight q subscript 2 over denominator left parenthesis 5 cross times 10 to the power of negative 2 end exponent right parenthesis squared end fraction space equals space 8 cross times 10 to the power of negative 3 end exponent cross times 9.8
On simplificaton,
                           straight q subscript 2 space equals space 4.356 space cross times space 10 to the power of negative 7 end exponent straight C

Let m be the mass of the upper ball. Let q1 represent the charge on
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A particle of mass m carrying charge + q1 is revolving around a fixed charge – q2 in a circular path of radius r. Calculate the period of revolution.

Since the particle carrying posive charge is revolving around another charge,

Electrostatic force = Centrifugal force

            14πε0q1 q2r2 = mrω2 =4π2mrT2

                  T2 = (4πε0) r2 (4π2 mr)q1 q2

                  T =4πrπε0 mrq1 q2

where T is the period of revolution.

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The electrostatic force between charges of 200 μC and 500 μC placed in free space is 5 gf. Find the distance between the two charges. Take g = 10 ms–2.

charge -q1 = 200 × 10-6C = 2 × 10-4C,charge -q2 = 500 × 10-6C = 5 × 10-4C, Electrostatic force -F = 5 gf = 5 × 10-3 kgf                                         = 5 × 10-3 × 10 N                                          = 5 × 10-2N,we have to find the distance between two charges i.e. r ?

Using the formula,  

                     F = 14πε0q1 q2r2, we get
       
             5 × 10-2 = 9×109×2×10-4×5×10-4r2
                
                r = 1.34 × 102 m
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