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3 years ago

12

An 800 V/m electric field is directed along the +x-axis. If the potential at x = 0 m is 2000 V, what is the potential at x = 2 m

?

1 answer:

Art [367]3 years ago

4 0

Answer:

400 V

Explanation:

Electric filed , E=800 V/m

$V_1=2000 V$

$x_1=0,x_2=2 m$

We know that

$E=\frac{V_1-V_2}{x_2-x_1}$

Using the formula

$800=\frac{2000-V_2}{2-0}$

$800=\frac{2000-V_2}{2}$

$2000-V_2=800\times 2=1600$

$V_2=2000-1600$

$V_2=400 V$

Hence, the potential at x=2 m=400 V

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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

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From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

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$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

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$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

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$t=20.7\text{ years}$

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