A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter cir

Question

A parallel-plate capacitor is constructed of two horizontal

12.0-cm-diameter circular plates. A 1.6 g plastic bead, with a charge of -5.2 nC is suspended between the two plates by the force of the electric field between them.

Lower. Constants It is impossible to tell. A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates A 1.6 g plastic bead, with a charge of -5.2 nC is suspended between the two plates by the force of the electric field between them. Previous Answe Correct Part B What is the charge on the positive plate? Express your answer to two significant figures and include the appropriate units. 0? Value Units Submit est Answer

Explanation / Answer

diameter (d) = 0.12 m which means radius (r) = 0.06 m
mass of bead = 1.6 g = 0.0016 kg
charge of bead = -5.2 nC = -5.2* 10^-9 C

Equation to use:
E = Q/(p*A), where p is permittivity constant which is 8.85 * 10^-12 C/(N*m^2), A is area of capacitor.
But you are solving for the charge on the positive plate (Q) so rearrange this formula:
Q = E*p*A

You know p since it s a constant. You can figure out the area (A) using area formula for a circle = pi(r)^2. But you don t have E. So we must solve for E using the formula E = F/q. The force needed to suspend the plastic bead can be found by F = mg, m is mass of the bead and g is gravity. q here is the charge of the bead.

Now you have E, p, and A, plug into Q = E*p*A

=> Q = (F/q)*(8.85 * 10^-12)*pi*(0.06)^2
=> Q = (mg/q)*(8.85 * 10^-12)*pi*(0.06)^2
=> Q = ((0.0016*9.8)/(5.2* 10^-9))*(8.85 * 10^-12)*pi*(0.06)^2
= 3*10^-7 C

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