In moving from A to B along an electric field line, the electric field does 6.4 × 10-19 J of work on an electron. If φ1, φ2 are equipotential surfaces, then the potential difference (VC-VA) is

Question

In moving from A to B along an electric field line, the electric field does 6.4 × 10-19 J of work on an electron. If φ1, φ2 are equipotential surfaces, then the potential difference (VC-VA) is (A) -4 V (B) 4 V (C) zero (D) 64 V

Tardigrade Admin · Accepted Answer

Work done by the field, W=q(-d V)=-e(VA-VB) =e(VB-VA)=e(VC-VA) (∵ VB=VC) ⇒ (VC-VA)=(W/e) =(6.4 × 10-19/1.6 × 10-19)=4 V

Q. In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times 1 0^{- 19} J$ of work on an electron. If $ϕ_{1}, ϕ_{2}$ are equipotential surfaces, then the potential difference $(V_{C} - V_{A})$ is

-4 V

4 V

zero

64 V

Work done by the field,
$W = q (- d V) = - e (V_{A} - V_{B})$
$= e (V_{B} - V_{A}) = e (V_{C} - V_{A})$
$(∵ V_{B} = V_{C})$
$\Rightarrow (V_{C} - V_{A}) = \frac{W}{e}$
$= \frac{6.4 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 4 V$

Q. In moving from A to B along an electric field line, the electric field does 6.4×10−19J of work on an electron. If ϕ1​,ϕ2​ are equipotential surfaces, then the potential difference (VC​−VA​) is

-4 V

4 V

zero

64 V

Work done by the field, W=q(−dV)=−e(VA​−VB​) =e(VB​−VA​)=e(VC​−VA​) (∵VB​=VC​) ⇒(VC​−VA​)=eW​ =1.6×10−196.4×10−19​=4V

Q. In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times 1 0^{- 19} J$ of work on an electron. If $ϕ_{1}, ϕ_{2}$ are equipotential surfaces, then the potential difference $(V_{C} - V_{A})$ is

Work done by the field,
$W = q (- d V) = - e (V_{A} - V_{B})$
$= e (V_{B} - V_{A}) = e (V_{C} - V_{A})$
$(∵ V_{B} = V_{C})$
$\Rightarrow (V_{C} - V_{A}) = \frac{W}{e}$
$= \frac{6.4 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 4 V$