Math, asked by sarikagoyal12ab, 10 days ago

Let the surface charge density of all three spheres be equal to sigma .
Now sphere of radius R has charge Q1 .
Since
Surface charge density = Charge / Surface area
So
\begin{gathered} \sigma = \frac{ Q_1}{4\pi R {}^{2} } \\ \\ \implies \: Q_1 = 4\pi\sigma R {}^{2} \end{gathered}σ=4πR2Q1​​⟹Q1​=4πσR2​
Since 4π is constant and sigma is constant for all three spheres so
Q \propto \: R {}^{2}Q∝R2
So we can conclude that
\begin{gathered}Q_1:Q_2:Q_3 \\ = R {}^{2} : (3R ) {}^{2} :(5R) {}^{2} \\ \\ \implies \: Q_1:Q_2:Q_3 \\ = R {}^{2} : 9R {}^{2} :25R {}^{2} \\ = 1 : 9 : 25\end{gathered}Q1​:Q2​:Q3​=R2:(3R)2:(5R)2⟹Q1​:Q2​:Q3​=R2:9R2:25R2=1:9:25​

is it true​

Answers

Answered by kunhimoideeneduvamal
0

Let the surface charge density of all three spheres be equal to sigma .

Now sphere of radius R has charge Q1 .

Since

Surface charge density = Charge / Surface area

So

\begin{gathered} \sigma = \frac{ Q_1}{4\pi R {}^{2} } \\ \\ \implies \: Q_1 = 4\pi\sigma R {}^{2} \end{gathered}σ=4πR2Q1⟹Q1=4πσR2

Since 4π is constant and sigma is constant for all three spheres so

Q \propto \: R {}^{2}Q∝R2

So we can conclude that

\begin{gathered}Q_1:Q_2:Q_3 \\ = R {}^{2} : (3R ) {}^{2} :(5R) {}^{2} \\ \\ \implies \: Q_1:Q_2:Q_3 \\ = R {}^{2} : 9R {}^{2} :25R {}^{2} \\ = 1 : 9 : 25\end{gathered}Q1:Q2:Q3=R2:(3R)2:(5R)2⟹Q1:Q2:Q3=R2:9R2:25R2=1:9:25

is it true

Answered by CarryminatiGiRl
0

Answer:

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Yes it is true!!

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