Question

Three charges all q=10 are placed at the edge of an equilateral triangle of side 2m . find the net potential energy of the system

Answer 1

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

$\green{\tt{\therefore{P.E_{Net}=1.35\:J}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Three \: charges = 10 \mu \: C \: \: \: \: (each) \\ \\ \tt: \implies Side \: of \: equilateral \: triangle = 2 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies P.E_{Net} = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies No. \: of \: pairs = \frac{n(n - 1)}{2} \\ \\ \tt: \implies No. \: of \: pairs = \frac{3 \times (3 - 1)}{2} \\ \\ \green{\tt: \implies No. \: of \: pairs = 3}\\ \\ \bold{As \: we \: know \: that} \\ \tt:\implies P.E_{Net} = \frac{k q_{1}q_{2}}{r_{12} } + \frac{kq_{2}q_{3}}{ r_{23}} + \frac{kq_{1}q_{3}}{ r_{13} } \\ \\ \tt:\implies PE._{Net} = \frac{k \times q \times q}{2} + \frac{k \times q \times q}{2} + \frac{k \times q \times q}{2} \\ \\ \tt:\implies P.E_{Net} = \frac{k {q}^{2} }{2} \times 3 \\ \\ \tt:\implies P.E_{Net} = \frac{3k {q}^{2} }{2} \\ \\ \tt:\implies P.E_{Net} = \frac{3 \times 9 \times {10}^{9} \times 100 \times {10}^{ - 12} }{2} \\ \\ \tt:\implies P.E_{Net} = \frac{27 \times {10}^{9 - 10} }{2} \\ \\ \tt:\implies P.E_{Net} = \frac{27 \times {10}^{ - 1} }{2} \\ \\ \green{\tt:\implies P.E_{Net} =1.35 \: J}$

Answer 2

Answer:

• The Potential Energy is 1.35 × 10¹² J

Given:

• There are three Charges of 10 C.
• Side of triangle (r) = 2 m.

Explanation:

$\rule{300}{1.5}$

Firstly finding the pairs of charges contributing to potential energy.and, for finding we know,

⇒ N = n (n - 1) / 2

Here,

• N Denotes no. of pairs of charge.
• n Denotes given amount of charge.

Substituting the values,

⇒ N = 3 × (3 - 1)/2

⇒ N = 3 × 2 / 2

⇒ N = 3

∴ There will be three pairs of charges contributing to the potential energy of the system.

They will be,

• Pair of q₁ and q₂
• Pair of q₁ and q₃  and,
• Pair of q₂ and q₃

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6)\put(2,2){\line(1,1){3}}\put(2,2){\line(1,0){6}}\put(8,2){\line(-1,1){3}}\put(1.7,1.8){ \sf q_2 }\put(8.1,1.8){ \sf q_3 }\put(4.9,5.2){ \sf q_1 }\put(4.5,1){\vector(-1,0){2.5}}\put(5.5,1){\vector(1,0){2.5}}\put(4.75,1){2 m}\put(7,-1){ \Bigg[\sf q_{1}=q_{2}=q_{3}= q=10\;C\Bigg] }\end{picture}$

Now, Let's find the potential energy,

$\displaystyle\longrightarrow\sf U=\dfrac{1}{4\;\pi\epsilon_{o}}\times \Bigg\lgroup \dfrac{q_{1}\;q_{2}}{r}+\dfrac{q_{1}\;q_{3}}{r}+\dfrac{q_{2}\;q_{3}}{r}\Bigg\rgroup\\\\\\\longrightarrow\sf U=\dfrac{1}{4\;\pi\epsilon_{o}}\times \Bigg\lgroup \dfrac{q^{2}}{r}+\dfrac{q^{2}}{r}+\dfrac{q^{2}}{r}\Bigg\rgroup\ \ \ \because\bigg[q_{1}=q_{2}=q_{3}\bigg]\\\\\\\longrightarrow\sf U=\dfrac{1}{4\;\pi\epsilon_{o}}\times3 \Bigg\lgroup \dfrac{q^{2}}{r}\Bigg\rgroup$

$\longrightarrow\sf U=9\times 10^{9}\times 3\times \dfrac{(10)^{2}}{2}\\\\\\\longrightarrow\sf U=27\times 10^{9}\times\dfrac{100}{2}\\\\\\\longrightarrow\sf U=27\times 10^{9}\times 50\\\\\\\longrightarrow\sf U=1.35\times 10^{12}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf U=1.35\times 10^{12}\;J}}}}$

∴ The Potential Energy is 1.35 × 10¹² J.

$\rule{300}{1.5}$