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A solid sphere has a charge Q Q distributed in its volume with a charge density d = k r a d=kr^a where k k and a a are constants and r r is distance from center. If electric field at r = R 2 r=\frac R2 is 1 8 \frac 18 times stronger than that of at r = R r=R , find a a .

Source: IIT JEE 2009.


The answer is 2.

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1 solution

Diego Barreto
May 8, 2015

1st part: By Gauss's Law: E d A = Q ε \oint { \vec { E } \cdot \vec { dA } } =\frac { Q }{ \varepsilon }

At the point r=R/2: E 1 π ( R 2 ) 2 = Q 1 ε { E }_{ 1 }\pi \left( \frac { R }{ 2 } \right) ^{ 2 }=\frac { { Q }_{ 1 } }{ \varepsilon }

At the point r=R: E 2 π R 2 = Q 2 ε { E }_{ 2 }\pi { R }^{ 2 }=\frac { { Q }_{ 2 } }{ \varepsilon }

We know that E2=8*E1, dividing the equations we get that: 32 Q 1 = Q 2 \boxed{32Q1=Q2}

2nd part (to avoid confusion, density will be "D", and derivatives "d"): D = K r a = d Q d V d Q = K r a d V D=K{ r }^{ a }=\frac { dQ }{ dV } \rightarrow \int { dQ } =K\int { { r }^{ a }dV }

V = 4 3 π r 3 d V d r = 4 π r 2 V=\frac { 4 }{ 3 } \pi { r }^{ 3 }\rightarrow \frac { dV }{ dr } =4\pi { r }^{ 2 }

d Q = K 4 π r a + 2 d r Q = 4 π K a + 3 r a + 3 \int { dQ } =K4\pi \int { { r }^{ a+2 }dr } \rightarrow Q=\frac { 4\pi K }{ a+3 } { r }^{ a+3 }

32 Q 1 = Q 2 32 4 π K a + 3 R a + 3 2 a + 3 = 4 π K a + 3 R a + 3 32 = 2 a + 3 32{ Q }_{ 1 }={ Q }_{ 2 }\rightarrow 32\frac { 4\pi K }{ a+3 } \frac { { R }^{ a+3 } }{ { 2 }^{ a+3 } } =\frac { 4\pi K }{ a+3 } { { R }^{ a+3 } }\rightarrow 32={ 2 }^{ a+3 }

Therefore, a = 2 \boxed{a=2}

I don't know if my solution is the easiest way of doing it. I think it's not considering this is a level 2 problem and my solution is not so simple. If you have a simpler solution please post it.

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