Question 407.
A small object with a mass of 270 mg carries a charge of 35.0 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 2.00 cm. If the thread makes an angle of 14.0° with the vertical, what is the potential difference between the plates?
Answer 407.
Let T = tension in the thread
and F = electrostatic force on the charge
For static equilibrium, balancing forces in the horizontal and vertical directions,
mg = Tcos14° and
F = Tsin14°
Taking the ratio,
F = mgtan14° ... ( 1 )
If V = potential difference between the plates,
electric field between the plates
= V/d = V/0.02 N/C
=> force on the charge
F = 35.0 x 10^-9 V/(0.02) ... ( 2 )
From ( 1 ) and ( 2 ),
35.0 x 10^-9 V/(0.02) = mgtan14°
=> Potential difference between the plates, V
= (0.02mgtan14°) / (35.0 x 10^-9)
= (0.02 * 270 x 10^-6 * 9.81 * tan14°) / (35.0 x 10^-9) volt
= 377.4 volt.
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A small object with a mass of 270 mg carries a charge of 35.0 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 2.00 cm. If the thread makes an angle of 14.0° with the vertical, what is the potential difference between the plates?
Answer 407.
Let T = tension in the thread
and F = electrostatic force on the charge
For static equilibrium, balancing forces in the horizontal and vertical directions,
mg = Tcos14° and
F = Tsin14°
Taking the ratio,
F = mgtan14° ... ( 1 )
If V = potential difference between the plates,
electric field between the plates
= V/d = V/0.02 N/C
=> force on the charge
F = 35.0 x 10^-9 V/(0.02) ... ( 2 )
From ( 1 ) and ( 2 ),
35.0 x 10^-9 V/(0.02) = mgtan14°
=> Potential difference between the plates, V
= (0.02mgtan14°) / (35.0 x 10^-9)
= (0.02 * 270 x 10^-6 * 9.81 * tan14°) / (35.0 x 10^-9) volt
= 377.4 volt.
Link to YA!
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