Engineers discover that the electric potential between two electrodes can be modeled as V(x)=V0ln(1+x/d), where V0 is a constant

Question

3 years ago

3

, x is the distance from the first electrode in the direction of the second, and dd is the distance between the electrodes.

1 answer:

Mashcka [7] · Answer 1 · 2022-04-17T21:14:06+03:00

Electric field strength halfway between the electrodes: $\frac{2V_0}{3d}$

Explanation:

The electric potential between the two electrodes is

$V(x)=V_0 ln(1+\frac{x}{d})$

where

$V_0$ is a constant

x is the distance from the first electrode

d is the distance between the electrodes

The relationship between electric field and electric potential is

$E=-\frac{dV}{dx}$

which means that the electric field is the derivative of the electric potential.

Therefore, calculating the derivative of V(x),

$E=-\frac{d}{dx}(V_0 ln(1+\frac{x}{d}))=-V_0 \frac{1/d}{1+\frac{x}{d}}$

Therefore, the electric field strength (so we neglect the negative sign) midway between the electrodes, at

$x=\frac{d}{2}$

is:

$E=V_0 \frac{1/d}{1+\frac{d/2}{d}}=V_0 \frac{1/d}{1+\frac{1}{2}}=V_0 \frac{1/d}{3/2}=\frac{2V_0}{3d}$

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