Question

Two point charges of magnitudes qı = 2.16 °C and q2 = 85.3 nC are 11.7 cm apart.
Find the magnitude of the force on each charge. ​

Answer 1

Answer:

apply colombs law

olombs law can easily applied on point charges

it states that firce between two point charges is directly proportianal to magnitude if product of charges and invesly to square if distance betwwen them.

see photo attacged for explanation

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Question :}}}}}$

Two point charges of magnitudes $\rm{Q_1 \: = \: 2.16 \: C}$ and $\rm{Q_2 \: = \: 85.3 \: C}$ are 11.7 cm apart. Find the magnitude of the force on each charge.

$\rule{200}{2}$

$\huge{\underline{\underline{\blue{\mathfrak{AnSwEr :}}}}}$

$\large \sf Given \begin{cases} \tt{Q_1 \: = \: 2.16 \: C} \\ \tt{Q_2 \: = \: 85.3 \: C} \\ \tt{Distance \: (d) \: = \: 11.7 \: cm } \\ \tt{Force \: = \: ?} \end{cases}$

Solution :

Convert Distance to m

$\sf{Distance \: (d) \: = \: 11.7 \: cm \: = \: 11.7 \: \times \: 10^{-2} \: m}$

$\rule{200}{2}$

Use Coulomb's Law :

$\huge{\boxed{\boxed{\rm{F \: = \: k \dfrac{Q_1 Q_2}{d^2}}}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 10^9 \: \times \: 2.16 \: \times \: 85.3}{(11.7 \: \times \: 10^{-2})^2}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 10^9 \: \times \: 184.24}{136.89 \: \times \: 10^{-4}}}} \\ \\ \implies {\sf{F \: = \: \dfrac{1658.23 \: \times \: 10^9}{136.89 \: \times \: 10^{-4}}}} \\ \\ \implies {\sf{F \: = \: 12.1 \: \times \: 10^9 \: \times \: 10^4}} \\ \\ \implies{\sf{F \: = \: 12.1 \: \times \: 10^{13}}} \\ \\ {\underline{\sf{\therefore \: \: Magnitude \: of \: Force \: is \: 12.1 \: \times \: 10^{13} \: N}}}$