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NEET-XII-Physics
31: Capacitors
- #6The energy density in the electric field created by a point charge falls off with the distance from the point charge as
(a)
1r
(b)
1r2
(c)
1r3
(d)
1r4.digAnsr: dAns : (d) `` \frac{1}{{r}^{4}}``
Energy density U is given by
U = `` \frac{1}{2}{\in }_{0}{E}^{2}`` ...(1)
The electric field created by a point charge at a distance r is given by
E = `` \frac{q}{4\,\mathrm{\,\pi \,}{\in }_{0}{r}^{\mathit{2}}}``
On putting the above form of E in eq. 1, we get
U = `` \frac{1}{2}{\in }_{0}{\left(\frac{q}{4\,\mathrm{\,\pi \,}{\in }_{0}{r}^{\mathit{2}}}\right)}^{2}``
Thus, U is directly proportional to `` \frac{1}{{r}^{4}}``.
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