Question

Charges of +10 μC, +20 μC and -20 μC are placed in air at the corners of an equilateral triangle having each side equal to 2 cm. Determine the force on charge +10 μC.

Answer 1

Answer:

• The Force on charge + 10 μC (F') is 4500 N.

Given:

1. Side of triangle (l) = 2 cm = 2 × 10⁻² m
2. Given Charges + 10 μC , + 20 μC, - 20 μC

Explanation:

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# Refer the attachment for the figure.

Here, we can clearly see that there will be Electrostatic force of repulsion between + 10 μC & + 20 μC (Like charges Repel).

and, there will be electrostatic force of attraction between + 10 μC & - 20 μC (unlike charges attract each other).

Now,

Let the electrostatic force of Repulsion between + 10μC & + 20μC be F₁.

Let the electrostatic force of attraction between + 10μC & - 20μC be F₂.

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Now, Finding the Electrostatic force of repulsion (F₁) + 10 μC and + 20 μC

From the formula,

$\large \; \bigstar\;\boxed{\tt F_1 = \dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{Q_1\;Q_2}{r^2}}$

Converting the charges into S.I units,

Q₁ = + 10 μC = + 10 × 10⁻⁶ C

Q₂ = + 20 μC = + 20 × 10⁻⁶ C

Substituting the values,

$\displaystyle \dashrightarrow\tt F_1=\dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{10\times10^{-6}\times 20\times10^{-6} }{(2\times 10^{-2})^2}\\\\\\\dashrightarrow\tt F_1=9\times 10^9\;.\;\dfrac{10\times 20\times 10^{(-6-6)}}{4\times 10^{-4}}\\\\\\\dag \quad \tt \dfrac{1}{4 \pi \epsilon_0}=9\times10^9\\\\\\\dashrightarrow\tt F_1=9\times 10^9\;.\;\dfrac{200\times 10^{-12}}{4\times 10^{-4}}\\\\\\\dashrightarrow\tt F_1=9\times 10^9\;.\; 50\times 10^{-12}\times 10^4$

$\displaystyle\dashrightarrow\tt F_1=9\times 10^9\times 50\times10^{-8}\\\\\\\dashrightarrow\tt F_1= 450\times 10^{9-8}\\\\\\\dashrightarrow\tt F_1=450\times 10\\\\\\\dashrightarrow \large \underline{\underline{\tt F_1 = 4500\;N}}$

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Now, Finding the Electrostatic force of attraction (F₂) + 10 μC and - 20 μC

From the formula,

$\large \; \bigstar\;\boxed{\tt F_2 = \dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{Q_1\;Q_2}{r^2}}$

Note, we have to take the magnitude of the charges i.e. two charges will be |Q₁| = 10μC and, |Q₂| = 20μC

Converting the charges into S.I units,

Q₁ = 10 μC = 10 × 10⁻⁶ C

Q₂ = 20 μC = 20 × 10⁻⁶ C

Substituting the values,

$\displaystyle \dashrightarrow\tt F_2=\dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{10\times10^{-6}\times 20\times10^{-6} }{(2\times 10^{-2})^2}\\\\\\\dashrightarrow\tt F_2=9\times 10^9\;.\;\dfrac{10\times 20\times 10^{(-6-6)}}{4\times 10^{-4}}\\\\\\\dag \quad \tt \dfrac{1}{4 \pi \epsilon_0}=9\times10^9\\\\\\\dashrightarrow\tt F_2=9\times 10^9\;.\;\dfrac{200\times 10^{-12}}{4\times 10^{-4}}\\\\\\\dashrightarrow\tt F_2=9\times 10^9\;.\; 50\times 10^{-12}\times 10^4$

$\displaystyle\dashrightarrow\tt F_2=9\times 10^9\times 50\times10^{-8}\\\\\\\dashrightarrow\tt F_2= 450\times 10^{9-8}\\\\\\\dashrightarrow\tt F_2=450\times 10\\\\\\\dashrightarrow \large \underline{\underline{\tt F_2 = 4500\;N}}$

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Since we can make out ,

⇒ F₁ = F₂

Therefore,

⇒ F₁ = F₂ = F

Both Forces F₁ & F₂ are inclined at 120° (180 - 60)° angle

From the formula of vector addition we know,

⇒ F' = 2F cos (θ/2)

(Where F' is the resultant)

Substituting the values,

⇒ F' = 2F × cos (120/2)

⇒ F' = 2F × cos(60)

⇒ F' = 2F × 1 / 2

⇒ F' = F

Therefore, the Resultant force is equal to the Magnitude of the force acting on the charge,

⇒ F' = 4500

⇒ F' = 4500 N

∴ The Force on charge + 10 μC (F') is 4500 N.

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