A charged ball on a string between two charged plates

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SUMMARY

A charged ball with a mass of 0.2 grams and a charge of 6E-9 Coulombs is suspended between two plates that are 5 cm apart, making a 30-degree angle with the vertical. The voltage difference across the plates was calculated using the electric field formula, resulting in a value of 96 volts, which was initially underestimated. The correct approach involves using the gravitational force and the tangent of the angle to determine the electric field, leading to a more accurate voltage calculation of 9440 volts.

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  • Knowledge of trigonometric functions, specifically tangent
  • Familiarity with basic physics equations related to electric charge and voltage
  • Ability to perform unit conversions and calculations in physics
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  • Study the relationship between electric field strength and voltage using the formula E = V/S
  • Learn about the forces acting on charged particles in electric fields
  • Explore the implications of angle measurements in physics problems involving forces
  • Review the concepts of gravitational force and its impact on charged objects
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Karol
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Homework Statement


A charged ball of mass 0.2[gr] and q=6E-9[Coulomb] hangs on a string between 2 plates that are 5[cm] apart. the ball makes 300 with the vertical. what is the voltage difference of the plates.

Homework Equations


Electric field: $$E[\frac{N}{Coulomb}]=\frac{V}{S[m]}$$
The force: $$F=E\cdot q$$

The Attempt at a Solution


The horizontal force: $$F_x=mg\cdot \tan 30^0=2E-4[Kg]\cdot 10\cdot \tan 30^0=11E-6[N]$$
The field: $$11E-6[N]=E\cdot 6E-9[Coulomb]\rightarrow E=1924.5[V]$$
Seems small.
##E=\frac{V}{0.05[m]}\rightarrow 1924.5=\frac{V}{0.05[m]} \rightarrow V=96[V]##
It should be 9440[V]
 
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Karol said:

Homework Statement


A charged ball of mass 0.2[gr] and q=6E-9[Coulomb] hangs on a string between 2 plates that are 5[cm] apart. the ball makes 300 with the vertical. what is the voltage difference of the plates.

Homework Equations


Electric field: $$E[\frac{N}{Coulomb}]=\frac{V}{S[m]}$$
The force: $$F=E\cdot q$$

What is the direction of the electric force?

Karol said:

The Attempt at a Solution


The horizontal force: $$F_x=mg\cdot \tan 30^0=2E-4[Kg]\cdot 10\cdot \tan 30^0=11E-6[N]$$

You mean $$F_x=mg\cdot \tan 30^0=2\cdot 10^{-4}\cdot 10\cdot \tan 30^0=11\cdot10^{-6}[N]$$
Use g=9.8 m/s2, and do not round off so much. And check the magnitude of the force.
 
Thanks, i solved
 

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