A uniform electric field is increasing at a constant rate overtime. If the displ

Question

A uniform electric field is increasing at a constant rate overtime. If the displacement current through 0.02m^2 surfacearea perpendicular to the electric field vector is 3 mA, find therate at which the electric field changes.
I used the formula that displacement current is Id= Eo * A * De/dt(rate), I = 3 x 10^-3 A E = 8.85 x 10 ^ -12 (electric constant) A = .02m^2
Plugging these in I come out to de/dt = 1.7 x 10^10 V / m*s or .017 V / m * Ts
Is this correct?
I used the formula that displacement current is Id= Eo * A * De/dt(rate), I = 3 x 10^-3 A E = 8.85 x 10 ^ -12 (electric constant) A = .02m^2
Plugging these in I come out to de/dt = 1.7 x 10^10 V / m*s or .017 V / m * Ts
Is this correct?

Explanation / Answer

I = 3 x 10^-3 A E = 8.85 x 10 ^ -12 N m^ 2/ C ^ 2 A = 0.02m^2 from this dE / dt = I d / [ o * A ]                         = 1.6949 * 10 ^ 10 V / m * s                         ~ 1.7 * 10 ^ 10 V / m*s

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