Question

A test charge q is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in Fig. (i) Calculate the potential difference between A and C (ii) At what point [of A and C] is the electric potential more and why?

Answer 1

P.D. does not depend upon the path alongwith the test charge q moves
`therefore" "E=(-dv)/(dr)=-((V_(C)-V_(A))/(d))-(V_(A)-V_(C))/(d)`
`d_(AC)=4" So, "V_(A)-V_(C)-Exx4-4^^(E )`
At point C, electric potential will be more since as potential decreases in the direction of electric field
`therefore" "V_(A)-E_(0)=E_(0)xxd.`