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 the figure- -i- Calculate the potential difference between A and C- -ii- At which point -of the two- is the electric potential and why-

Question

Toppr Admin · Accepted Answer

-i-Electric field is given by-x2192-E-E-i-x2212-dVdx-iHence- VC-x2212-VA-x2212-x222B-Edx-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-VA-x2212-VC-x2212-4E-ii-Electric potential is more at point C as electric field is directed from higher potential region to lower potential region-

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 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?

(i)Electric field is given by:→E=E^i=−dVdx^iHence, VC−VA=−∫Edx VA−VC=−4E(ii)Electric potential is more at point C as electric field is directed from higher potential region to lower potential region.

(i)
Electric field is given by:
$\to E = E^i = - \frac{d V}{d x}^i$
Hence, $V_{C} - V_{A} = - \int E d x$
$V_{A} - V_{C} = - 4 E$

(ii)
Electric potential is more at point C as electric field is directed from higher potential region to lower potential region.