How Much Energy Is Required to Move a Charge to Infinity?

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SUMMARY

This discussion addresses the calculation of electrical potential energy and electric potential for various charge configurations. For Problem 10, the electrical potential energy expended in moving a +8.6 µC charge to infinity is determined to be 0 J, as the potential approaches zero at infinity. In Problem 11, the electric potential at the midpoint of the base of an isosceles triangle with charges of 3.9 x 10^-9 C is calculated to be 10.46 V. Problem 18 calculates the electrical potential energy stored in a capacitor with a capacitance of 0.25 µF operating at 6300 V, yielding a result of 496.125 J.

PREREQUISITES
  • Coulomb's Law for electric potential energy calculations
  • Understanding of electric potential and its formula V = k_e * (q/r)
  • Knowledge of capacitor energy storage formula U = (1/2) * C * V^2
  • Basic geometry for calculating distances in charge configurations
NEXT STEPS
  • Study Coulomb's Law and its applications in electrostatics
  • Learn about electric potential energy in different charge configurations
  • Explore the principles of capacitors and their energy storage capabilities
  • Investigate the effects of distance on electric potential in various geometries
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Students and professionals in physics, electrical engineering, and anyone involved in electrostatics or capacitor design will benefit from this discussion.

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Problem 10.
Given: k_e=8.98755*10^9 nm^2/C^2 and g=9.8 m/s^2.
Three charges are at three corners of a rectangle a charge of +8.6uC is at the upper left hand corner, a +2.5uC is at the lower left hand corner, and a +3.7 uC is ath the lower right hand corner. The length of the rectangle is 5.1 cm and the width is 3.1 cm.
How much electrical potential energy would be expended in moving the 8.6 uC charge to infinity? Answer in units of J.

Problem 11.
Given: k_e=8.98755*10^9 nm^2/C^2 and g=9.8 m/s^2.
Three charges are located at the vertices of an isosceles triangle. A positive charge is
a the top vertice, and two negative charges are at the bases. The length of the base is 2.7 cm and the other two sides is 5.4 cm.
Calculate the electric potential at the midpoint of the base if the magnitude of each charge is 3.9*10^-9C. Answer in units of V.
note: What formula would you use?

Problem 18.
A parallel-plate capacitor has a capacitance of 0.25 uF and is to be operated at 6300 V.
What is the electrial potential energy stored in the capacitory at the operating potential difference? Answer in units of J.
Note: I don't know where to start.
 
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Calculate the potential(V) at the point where a charge of 8.6uC is located due to the other two charges. At infinity the potential due to other charges will be zero

Therefore Electric Potential Energy is = q(V-0)

2) Potential at a point due to charge q is kq/r where r is the distance b/w the charge and the point where potential is to be calculated

3) apply the formula 1/2 CV^2
 


For Problem 10:

To solve this problem, we can use the formula for electrical potential energy: U = k_e * (q1*q2)/r, where k_e is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.

First, we need to find the distance between the +8.6uC charge and infinity. Since the charge is moving to infinity, we can assume that the distance is very large, approaching infinity. Therefore, we can use the limit as r approaches infinity, which is equal to zero.

Plugging in the given values, we get: U = (8.98755*10^9 nm^2/C^2) * (8.6*10^-6 C * 0 C)/0 = 0 J.

Therefore, no electrical potential energy would be expended in moving the +8.6uC charge to infinity.

For Problem 11:

To solve this problem, we can use the formula for electric potential: V = k_e * (q/r), where k_e is the Coulomb's constant, q is the charge, and r is the distance between the charge and the point where we want to find the potential.

In this case, we need to find the potential at the midpoint of the base, which is half the distance of the base. Therefore, r = 1.35 cm.

Plugging in the given values, we get: V = (8.98755*10^9 nm^2/C^2) * (3.9*10^-9 C/1.35 cm) = 10.46 V.

The formula used for this problem is the formula for electric potential.

For Problem 18:

To solve this problem, we can use the formula for electric potential energy stored in a capacitor: U = (1/2)*C*V^2, where C is the capacitance and V is the potential difference.

Plugging in the given values, we get: U = (1/2)*(0.25*10^-6 uF)*(6300 V)^2 = 496.125 J.

Therefore, the electrical potential energy stored in the capacitor at the operating potential difference is 496.125 J. The formula used for this problem is the formula for electric potential energy stored in a capacitor.
 

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