Question

Find the magnitude of the electric force between the charges 0.12 C and 0.33 C at a separation of 2.5 m. k=8.99×109N⋅m2/C2.

Answer 1

Solution

• Force experienced by the charges is 5.7 × 10^7 N

Given

• Distance of seperation (r) = 2.5 m

• Charges (Q,q) = (0.12,0.33)

• K = 8.9 × 10^9 N m²/C² ≈ 9 × 10^9 Nm²/C²

To finD

Electric Force between the system of charges

Using the Relation,

$\sf F = \dfrac{1}{4 \pi \epsilon} \times \dfrac{Qq}{r^2} \\ \\ \longrightarrow \boxed{\boxed{\sf F = \dfrac{K Q q}{r^2}}}$

Substituting the values,we get :

$\sf F = 9 \times 10^9 \times \dfrac{0.12 \times 0.33}{(2.5)^2} \\ \\ \dashrightarrow \sf F = \dfrac{9 \times 12 \times 33 \times 10^5}{625 \times 10^{-2}}\\ \\ \dashrightarrow \sf F = \dfrac{3564}{625} \times 10^{7} \\ \\ \dashrightarrow \sf F = 5.7024 \times 10^7 N \\ \\ \dashrightarrow \boxed{\boxed{\sf F = 5.7 \times 10^7 \ N}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

$\sf Given \begin{cases} \rm{First \: Charge \: (Q_1) \: = \: 0.12 \: C} \\ \rm{Second \: Charge \: (Q_2) \: = \: 0.33 \: C} \\ \rm{Distance \: between \: charges \: (d) \: = \: 2.5 \: m} \end{cases}$

Here, we have to find Magnitude of electrostatic force.

Use Coulomb's Law :

$\large{\boxed{\boxed{\rm{F \: = \: k \dfrac{Q_1 Q_2}{d^2}}}}} \\ \\ \implies {\sf{F \: = \: 9 \: \times \: 10^9 \: \times \: \dfrac{0.12 \: \times \: 0.33}{(2.5)^2}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 12 \: \times \: 33 \: \times \: 10^5}{6.25}}} \\ \\ \implies {\sf{F \: = \: \dfrac{3564 \: \times \: 10^5}{6.25}}} \\ \\ \implies {\sf{F \: = \: \dfrac{3564 \: \times \: 10^5}{625 \: \times \: 10^{-2}}}} \\ \\ \implies {\sf{F \: = \: \dfrac{3564 \: \times \: 10^7}{625}}} \\ \\ \implies {\sf{F \: = \: 5.70 \: \times \: 10^7}} \\ \\ \implies {\sf{F \: = \: 5.7 \: \times \: 10^7}} \\ \\ {\underline{\sf{\therefore \: Magnitude \: of \: Electrostatic \: Force \: is \: 5.7 \: \times \: 10^7 \: N}}}$