An electric field nbsp-E -8594-i -8594- Ax exists in space- where A - 10 V m minus-2- Take the potential at -10 m- 20 m- to be zero- Find the potential at the origin-

Given: Electric field intensity, E #8594;=i^Ax=10xi^ Potential, dV= E #8594;. #160;dx #8594;= 10xdx On integrating, we get V=10 #215;x22= 5x2100V=5 ...

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Standard XII

Physics

Simple Circuit

An electric f...

Question

An electric field $\vec{E} = \vec{i}$ Ax exists in space, where A = 10 V m⁻². Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.

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Solution

Given:
Electric field intensity, $\vec{E} = \hat{i} Ax = 10 x \hat{i}$
Potential, $d V = - \vec{E} . \vec{d x} = - 10 x d x$
On integrating, we get
$V = 10 \times \frac{x^{2}}{2} = - {[5 x^{2}]}_{10}^{0} V = 5 \times 100 = 500 V$
So, at the origin, the potential is 500 V.

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An electric field E→=i→ Ax exists in space, where A = 10 V m−2. Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.

Given: Electric field intensity, E→=i^Ax=10xi^ Potential, dV=-E→. dx→=-10xdx On integrating, we get V=10×x22=-5x2100V=5×100=500 V So, at the origin, the potential is 500 V.

An electric field $\vec{E} = \vec{i}$ Ax exists in space, where A = 10 V m⁻². Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.

Given:
Electric field intensity, $\vec{E} = \hat{i} Ax = 10 x \hat{i}$
Potential, $d V = - \vec{E} . \vec{d x} = - 10 x d x$
On integrating, we get
$V = 10 \times \frac{x^{2}}{2} = - {[5 x^{2}]}_{10}^{0} V = 5 \times 100 = 500 V$
So, at the origin, the potential is 500 V.