Question

3. Two charges of magnitude 10 units and 20
units are separated by certain distance. Now
both the charges are brought to contact and
again separated to initial position. What will
be the ratio of initial and final force?​

Answer 1

Answer:

Hey mate your answer is

Explanation:

lets say their centers are separated by r units. force between them is then

F1→=q1q2r2=−200r2r^

where - sign represents the fact that they are attracted to each other

since they are metals of same kind hence when touched, their charges will equalize and each one will have 10−202=−5units of charge.

now if they are moved back to their initial position, force between them

F2→=+25r2r^

where + sign represents the fact that they repel each other since they both have -ve charges.

ratio of magnitude of forces

F2F1=25200=18

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{F_{Initial}}{F_{Final}}=\frac{8}{9}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Two \: chagres = 10 \: \mu c \: and \: 20 \: \mu \: c \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \frac{F_{Initial} }{F_{Final} } =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F_{Initial} = \frac{k q_{1} q_{2}}{ {r}^{2} } \\ \\ \tt: \implies F_{Initial} = \frac{k \times 10 \times 20}{ {r}^{2} } \\ \\ \tt: \implies F_{Initial} = \frac{200k}{ {r}^{2} } - - - - - (1) \\ \\ \bold{For \: Final \: force : } \\ \\ \tt\circ\:Magnitude\:of\:charge=15\:\mu C\:\:\:\:\:\:(each)\\\\\tt: \implies F_{Final} = \frac{kq_{1} q_{2} }{ {r}^{2} } \\ \\ \tt: \implies F_{Final} = \frac{k \times 15 \times 15}{r^{2}} \\ \\ \tt: \implies F_{Final} =\frac{225k}{r^{2}} - - - - - (2) \\ \\ \bold{For \: Ratio : } \\ \tt: \implies \frac{F_{Initial} }{F_{Final}} = \frac{ \frac{200k}{ {r}^{2} } }{ \frac{225k}{r^{2} }} \\\\ \tt: \implies \frac{F_{Initial} }{F_{Final}} = \frac{200}{225} \\ \\ \green{\tt: \implies \frac{F_{Initial} }{F_{Final}} = \frac{8}{9}}$