Electric fields and equipotentials

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SUMMARY

The discussion focuses on calculating forces and electric fields involving test charges in the presence of other charges. For a test charge of q = -4 µC, the net force calculated is 190 N, with individual forces of 80 N to the left and 269.7 N to the right. The electric field at the position of the test charge is determined to be -4.75 x 10^7 N/C. Additionally, the work done in moving a charge of 2.5 µC along an equipotential at 9 V is calculated as -2.25 x 10^-5 J.

PREREQUISITES
  • Understanding of Coulomb's Law (F=ke|q1|q2|/r^2)
  • Knowledge of electric field calculations (E=F/q)
  • Familiarity with potential energy concepts in electrostatics
  • Basic algebra for manipulating equations and units
NEXT STEPS
  • Study the implications of negative charges in electric fields
  • Learn about equipotential surfaces and their properties
  • Explore the concept of work done in electric fields in greater detail
  • Investigate the relationship between electric field strength and potential difference
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone needing to solve problems related to electric forces and fields.

dasblack
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Homework Statement



The figure below shows a test charge q between the two positive charges. Find the force (in Newtons) on the test charge for q = -4 µC.
a)Give a positive answer if the force is to the right and a negative answer if the force is to the left.
wpl2.jpg


b.)For the previous question, find the electric field (in Newtons/coulomb) at the position of test charge. Again, supply a positive value if the electric field points to the right and a negative value if it points to the left.

Homework Equations


F=ke|q1|q2|/r^2
E=F/qo

The Attempt at a Solution


a)F21=ke(2x10-6)(4x10-6)/(3x10-2)^2 = 80 to the left
F23=ke(4x10-6)(3x10-6)/(2x10-2)^2 = 269.7 to the right
F2 = 190

b)E=190/-4x10-6 = -4.75e7

Homework Statement



How much work (in joules) is done in moving a charge of 2.5μC a distance of 22 cm along an equipotential at 9~V?

Homework Equations


W=-change in PE, where PE = q(change in volts)

The Attempt at a Solution



W=-2.5e-6(9)= -2.25e-5 J
 
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Hmm I'm not sure about the negative for qo
 

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