Question

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Answer 1

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value
`v_2 = 4 √(10) \ m/s`

Step-by-step explanation:

From the question we are told that

The charge on the first sphere is
`q_1 = 2\mu C = 2*10^(-6) \ C`

The charge on the second sphere is
`q_2 = 8 \mu C = 8*10^(-6) \ C`

The mass of the second charge is
`m = 1.50 \ g = 1.50 *10^(-3) \ kg`

The distance apart is
`d = 0.4 \ m`

The speed of the second sphere is
`v_1 = 20 \ ms^(-1)`

Generally the total energy possessed by when
`q_2` and
`q_1` are separated by
`0.8 \ m` is mathematically represented

`Q = KE + U`

Here KE is the kinetic energy which is mathematically represented as

`KE = (1 )/(2) m (v_1)^2`

substituting value

`KE = (1 )/(2) * ( 1.50 *10^(-3)) (20 )^2`

`KE = 0.3 \ J`

And U is the potential energy which is mathematically represented as

`U = (k * q_1 * q_2 )/(d )`

substituting values

`U = (9*10^9 * 2*10^(-6) * 8*10^(-6) )/(0.8 )`

`U = 0.18 \ J`

So

`Q = 0.3 + 0.18`

`Q = 0.48 \ J`

Generally the total energy possessed by when
`q_2` and
`q_1` are separated by
`0.4 \ m` is mathematically represented

`Q_f = KE_f + U_f`

Here
`KE_f` is the kinetic energy which is mathematically represented as

`KE_f = (1 )/(2) m (v_2^2`

substituting value

`KE_f = (1 )/(2) * ( 1.50 *10^(-3)) (v_2 )^2`

`KE_f = 7.50 *10^( -4) (v_2 )^2`

And
`U_f` is the potential energy which is mathematically represented as

`U_f = (k * q_1 * q_2 )/(d )`

substituting values

`U_f = (9*10^9 * 2*10^(-6) * 8*10^(-6) )/(0.4 )`

`U_f = 0.36 \ J`

From the law of energy conservation

`Q = Q_f`

So

`0.48 = 0.36 +(7.50 *10^(-4) v_2^2)`

`v_2 = 4 √(10) \ m/s`