Calculate the work done by electric field when a point charge q is moved from point B to A along the curved path. Given that electric field is created by the stationary charge Q and $V_{A} = 200 V, V_{B} = 100 V, q = 0.05 C$ , length of line segment AB = 10 cm, length of curved path = 20 cm.

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Given - -xA0- q-0-05C -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-VA-200V -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- VB-100VAs the electric field is an conservative field- thus the work done in moving a charge from one point to another is path independent-x2234- -xA0- WBA-q-VA-x2212-VB-0-05-200-x2212-100-5 -xA0-J

Calculate the work done by electric field when a point charge q is moved from point B to A along the curved path. Given that electric field is created by the stationary charge Q and VA=200V,VB=100V,q=0.05C, length of line segment AB = 10 cm, length of curved path = 20 cm.5 J15 J0 J20 J10 J

Given : q=0.05C VA=200V VB=100VAs the electric field is an conservative field, thus the work done in moving a charge from one point to another is path independent.∴ WBA=q(VA−VB)=0.05(200−100)=5 J

Given : $q = 0.05 C$ $V_{A} = 200 V$ $V_{B} = 100 V$
As the electric field is an conservative field, thus the work done in moving a charge from one point to another is path independent.
$∴$ $W_{B A} = q (V_{A} - V_{B}) = 0.05 (200 - 100) = 5$ J