Electric field strength halfway between the electrodes: [tex]\frac{2V_0}{3d}[/tex]
Explanation:
The electric potential between the two electrodes is
[tex]V(x)=V_0 ln(1+\frac{x}{d})[/tex]
where
[tex]V_0[/tex] 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
[tex]E=-\frac{dV}{dx}[/tex]
which means that the electric field is the derivative of the electric potential.
Therefore, calculating the derivative of V(x),
[tex]E=-\frac{d}{dx}(V_0 ln(1+\frac{x}{d}))=-V_0 \frac{1/d}{1+\frac{x}{d}}[/tex]
Therefore, the electric field strength (so we neglect the negative sign) midway between the electrodes, at
[tex]x=\frac{d}{2}[/tex]
is:
[tex]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}[/tex]
Learn more about electric fields:
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brainly.com/question/4273177
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