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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 shows in the figure. (i) Calculate the potential difference between A and C. (ii) At which point (of the two) is the electric potential more and why ? |
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Answer» Solution :(i) As electric field is a conservative field, work doene in MOVING the TEST charge q is exactly same along the path ABC or along straight path AC. Distance`AC = vec R = - 4 hati`, and force applied against the electric field `vecF_("ext")= - vecE_q= -E_qhati` `:.` Work done `W_(AC) = vec F_("ext").vecr = [-EQ hat i].[-4hati] = + 4 Eq` `:.` Potential difference between C and A `V_C - V_A= (W_(AC))/q = (4Eq)/q = + 4E` (ii) Electric potential at point C is GREATER i.e., `V_C gt V_A`. |
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