328. Coulomb’s law can be derived from Gauss law. State True/ False
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a Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
a
See lessExplanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
A Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2 .
A
See lessExplanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
.
Answer: a Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2 .
Answer: a
See lessExplanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
.
Answer: a Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2 .
Answer: a
See lessExplanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
.
Answer: a Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2 .
Answer: a
See lessExplanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
.
a) Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2 .
a)
See lessExplanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2
.