Physics, 25.01.2020 03:31 olorteguimikie

The electrostatic force of attraction between two charged objects separated by a distance of 1.0 cm is given by f. if the distance between the objects were increased to 5.0 cm, what would be the electrostatic force of attraction between them?

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Answer from: leianagaming

New force between them will become $\frac{1}{36}$ times

Explanation:

Let the charge on both the object are $q_1\ and\ q_2$ and distance between them is is given 1 cm

So r = 1 cm = 0.01 m

Electric force between them is given F

According to Coulomb's between two charges is given by

$F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}$

According to question $F=\frac{Kq_1q_2}{0.01^2}=\frac{Kq_1q_2}{10^{-4}}$ -----------eqn 1

Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m

So new force $F_{new}=\frac{Kq_1q_2}{0.06^2}=\frac{Kq_1q_2}{36\times 10^{-4}}$ ------------------eqn 2

Comparing eqn 1 and eqn 2

$F_{new}=\frac{F}{36}$

Answer from: Quest

. west

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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