How Does Changing Charge Affect Potential and Electric Field?

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SUMMARY

The discussion focuses on calculating the electric potential (V) and electric field (E) due to two point charges, q and -q, separated by a distance d. The potential at a point P, located a distance z above the center of the charges, is derived using the equation V = K ∑ q_i/r_i. When the right-hand charge is changed to -q, the potential remains 0, but the electric field, calculated as E = -∇V, does not equal 0, indicating a non-zero electric field despite the potential being zero.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with the equation V = K ∑ q_i/r_i
  • Knowledge of vector calculus, specifically gradient operations
  • Basic concepts of point charge interactions
NEXT STEPS
  • Study the implications of charge distributions on electric fields
  • Learn about the gradient operator in vector calculus
  • Explore the concept of superposition in electric potential
  • Investigate the relationship between electric potential and electric field strength
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in understanding the effects of charge distributions on electric potential and fields.

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Homework Statement


Find the potential at a distance z above the center of the charge distributions q and -q, which are distance d apart. Also, compute E= -[tex]\nabla[/tex]V and compare your answer. Suppose that we changed the right-hand charge to -q , what then is the potential at P? What field does that suggest?


Homework Equations


V = K [tex]\sum q_{}i/r_{}i[/tex]

The Attempt at a Solution


I calculated V without changing the right side charge to -q and V came out to be 0 but how do I verify E = -[tex]\nabla[/tex]V since, V is 0, but E won't be 0. I am confused =$.
 
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Someone else asked pretty much the same question here
 

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