M4: THE COULOMB BALNCE Name: Attach a copy of the Pre-Lab answers. Area of the p

Question

M4: THE COULOMB BALNCE Name: Attach a copy of the Pre-Lab answers. Area of the plates- Thickness of one plate d' 60 la m Thickness of the Plastic Spacer O Plate Separation d (hd' do) Part A: Varying the Voltage: Mass Removed 0 mg 10 mg 20 mg 30 mg 40 mg 50 mg 60 mg 70 mg (kg) Potential (volts) o 10 L50 25 90 10 12 Part B: Attach a copy of the graph of V2 versus m Slope of the best linear fit C2/Nm Your calculated value for eo C2/Nm Percent difference between accepted value- Conclusion Questions: The limiting feature of this experiment is the maximum voltage that the power supply can provide. We used a plate separation of 0.3 cm and found that the maximum mass which could be removed from the top plate was about 70 mg at maximum voltage. Suppose we have a plate separation of 0.6 cm instead. Would the maximum mass that could be removed increase or decrease? Gustify your answer with the equations you calculated in the pre-lab Using your values for A and d, find the charge on the positive plate when the voltage is 200 volts This experiment could have been preformed by keeping the masses constant and varying the distance d instead. How would you expect the voltage to depend on d? (Linearly, inversely, proportional to the square of d, or something else)? One of the main sources oferror in this experiment is the measurement of d, which had to be kept arbitrarily small in order to use reasonable (non-lethal!) voltages A 10% error in the measurement of d would cause what percentage error in your Eo calculation?

Explanation / Answer

Pre-lab for experiment M4:

part A:

E=q/(epsilon*A)=V/d

==>q=(epsilon*A/d)*V

part B:

force=E*q=V*q/(2*d)

=V*epsilon*A*V/(2*d^2)

=epsilon*A*V^2/(2*d^2)

part C:

now this force should balance the weight of the particle.

hence epsilon*A*V^2/(2*d^2)=m*g

==>V^2=(2*d^2*g/(epsilon*A))*m

so slope of the graph of V^2 vs m is equal to 2*d^2*g/(epsilon*A)

part D:

total capacitance of a parallel plate capacitor=epsilon*A/d

then total charge=capacitance*potential difference

=epsilon*A*V/d

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