Question

Two point charges, QA =+8 μC and QB =-5 μC are separated by a distance 10 cm. What is the magnitude of the electric force. The constant k = 8.988*109 Nm^2C^−2=9*10^9 Nm^2C^−2.

Answer 1

Answer:

See in the image.

Explanation:

See the answer in the image.

Answer 2

Answer:

• The magnitude of Electric force (F) is 36 N

Given:

1. First charge (Q_A) = + 8 μC = 8 × 10⁻⁶ C
2. Second Charge (Q_B) = - 5 μC = - 5 × 10⁻⁶
3. Distance separated(r) = 10 cm =  0.1 m

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\large \bigstar\; \boxed{\tt F = \dfrac{1}{4\pi \epsilon_0}\;.\; \dfrac{Q_A \; Q_B}{r^2}}$

$\frak{Here}\begin{cases}\sf{Q_A}\text{ Denotes First Charge}\\ \sf{Q_B}\text{ Denotes Second Charge}\\\text{r Denotes Distance between charges}\\\sf{\epsilon_0}\text{ Denotes Permittivity of Free Space}\end{cases}$

Now,

$\large \boxed{\tt F = \dfrac{1}{4\pi \epsilon_0}\;.\; \dfrac{Q_A \; Q_B}{r^2}}$

Substituting the values,

$\displaystyle \dashrightarrow\tt F=\dfrac{1}{4\pi\epsilon_0}\;.\;\dfrac{8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\\\\dag\quad\tt \dfrac{1}{4\pi\epsilon_0}=K\\\\\\\dashrightarrow\tt F=K\times\dfrac{40\times 10^{-6-6}}{0.01}\quad\because[K=9\times 10^9]\\\\\\\dashrightarrow\tt F=9\times 10^9\times 40\times 10^2\times10^{-12}\\\\\\\dashrightarrow\tt F=360\times10^{9-12}\\\\\\\dashrightarrow\tt F=36000\times10^{-3}\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\tt F=36\;N }}}}$

∴ The magnitude of Electric force (F) is 36 N.

Note:-

• The units of charges are changed ( μC to C ) to get the results in S.I units.
• The unit of Distance are changed ( cm to m ) to get the result in S.I units.
• The second charge's only magnitude is taken, So, that we can get the magnitude of the Force.

$\rule{300}{1.5}$