Question

how far from 0.60 microcoulomb charges would a point be if it had a value of 10 kv? if the point were to move four times the initial distance, how much potential change would occur? Is it increase or decrease?
(Please give the answer in detail and fully solved)​

Answer 1

Answer:0.54m

Explanation:

Given: q=0.60 ×10^-6 C

V= 10kV = 10^4 V

R=?

V=kQ/R

R= kQ/V = 9×10^9 × 0.60×10^-6/ 10^4

By calculating, we get,

R= 0.54m

Answer 2

Given:

0.60 micro-C charge has potential of 10 kV at a certain distance.

To find:

• Value of distance

• Change in Potential when distance is made 4 times

Calculation:

Let the required distance be r .

$\therefore \: 10000 = \dfrac{kq}{r}$

$= > 10000 = \dfrac{(9 \times {10}^{9} ) \times (0.6 \times {10}^{ - 6}) }{r}$

$= > \dfrac{ 36 \times {10}^{2} }{r} = 10000$

$= > \dfrac{36}{r} = 100$

$= > r = \dfrac{36}{100} = 0.36 \: m$

So the required distance will be 0.36 m.

Potential at distance r = 10 kV

New distance be 4r ;

$\rm{new \: potential = \dfrac{kq}{4r} }$

$= > \rm{new \: potential = \dfrac{1}{4} \times \bigg \{\dfrac{kq}{r} } \bigg \}$

$= > \rm{new \: potential = \dfrac{1}{4} \times \bigg \{ 10000\bigg \}}$

$= > \rm{new \: potential = 2500 \: volt}$

So change in Potential will be :

∆V = 2500 - 10000 = -7500 volt.

So , potential decreased with increase in distance.