Chapter 22: Q. 58 (page 626)

Two equal point charges 2.5 cm apart, both initially neutral, are being charged at the rate of 5.0 nC/s. At what rate (N/s) is the force between them increasing 1.0 s after charging begins?

Short Answer

Expert verified

The rate at which the force is increasing is $7.12 \times 10^{- 4} N / s$ .

Step by step solution

Given Information

The given point charges are equal

Separation of the two points $= r = 2.5 c m = 2.5 \times 10^{- 2} m$

Rate of change of charge $= \frac{d q}{d t} = 5.0 n C / s = 5.0 \times 10^{- 9} C / s$

Calculation

Since it is given that the charges are the same, the electrostatic force can be given as:

$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$

Where,

$\frac{1}{4 {πε}_{0}} = 8.9 \times 10^{9} N m^{2} / C^{2}$

By differentiating the above equation with respect to time, we get the rate of change of force as:

$\frac{d F}{d t} = \frac{1}{4 π ε_{0} r^{2}} 2 q \frac{d q}{d t} . . . . . (1)$

The charge of the particles after 1.0 s can be calculated as:

localid="1648461224815" $q = \frac{d q}{d t} t q = 5.0 \times 10^{- 9} \times 1.0 q = 5.0 \times 10^{- 9} C$

By substituting this value and the given values in equation (1), we get,

$\frac{d F}{d t} = \frac{(8.9 \times 10^{9})}{{(2.5 \times 10^{- 2})}^{2}} \times 2 \times (5.0 \times 10^{- 9}) (5.0 \times 10^{- 9}) \frac{d F}{d t} = 7.12 \times 10^{- 4} N / s$

Final answer

Hence, for the given conditions, the force is increasing at a rate of $7.12 \times 10^{- 4} N / s$ .

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Two equal point charges 2.5 cm apart, both initially neutral, are being charged at the rate of 5.0 nC/s. At what rate (N/s) is the force between them increasing 1.0 s after charging begins?

Short Answer

Step by step solution

Given Information

Calculation

Final answer

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