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Question: **ILLUSTRATION 3.2** A charge particle q=-10µC is carried along OP and PQ and then back to O along ...

ILLUSTRATION 3.2

A charge particle q=-10µC is carried along OP and PQ and then back to O along QO as shown in figure, in an electric field E‾\overline{E}=(x+2y)i^\hat{i}+2xj^\hat{j}. Find the work done by an external agent in (a) each path and (b) the round trip.

Answer
  • Path OP: +20 μJ+20\,\mu J
  • Path PQ: +80 μJ+80\,\mu J
  • Path QO: −100 μJ-100\,\mu J
  • Round Trip: 0 J0\,J
Explanation

Solution

We first note that the given electric field

E⃗=(x+2y)i^+2xj^\vec{E}=(x+2y)\hat{i}+2x\hat{j}

is conservative because

∂(x+2y)∂y=2and∂(2x)∂x=2.\frac{\partial (x+2y)}{\partial y} = 2 \quad \text{and} \quad \frac{\partial (2x)}{\partial x} = 2.

Thus, we can introduce a potential function V(x,y)V(x,y) such that

E⃗=−∇V.\vec{E}=-\nabla V.
  1. Find the potential function V(x,y)V(x,y):

From

Ex=−(∂V∂x)=x+2y⟹∂V∂x=−(x+2y).E_x = -( \frac{\partial V}{\partial x}) = x+2y \quad \Longrightarrow \quad \frac{\partial V}{\partial x} = -(x+2y).

Integrate with respect to xx:

V(x,y)=−x22−2xy+f(y).V(x,y) = -\frac{x^2}{2} -2xy + f(y).

Next, using

Ey=−∂V∂y=2x⟹∂V∂y=−2x.E_y= -\frac{\partial V}{\partial y} =2x \quad \Longrightarrow \quad \frac{\partial V}{\partial y} = -2x.

But from our expression,

∂V∂y=−2x+f′(y).\frac{\partial V}{\partial y} = -2x + f'(y).

Equating gives f′(y)=0f'(y)=0 so that f(y)f(y) is a constant. We may take the constant as zero. Thus,

V(x,y)=−x22−2xy.V(x,y) = -\frac{x^2}{2} -2xy.
  1. Work done by the external agent:

For slow (or quasi-static) displacement, the work done by the external agent is equal to the change in the potential energy of the charge. Since the potential energy is U=qVU=qV and q=−10 μC=−10×10−6 Cq=-10\,\mu C = -10\times10^{-6}\,C, we have:

Wext=ΔU=q (Vfinal−Vinitial).W_{\text{ext}} = \Delta U = q\,(V_{\rm final}- V_{\rm initial}).

  • Path OPOP: (from O(0,0)O(0,0) to P(2,0)P(2,0))

    V(0,0)=−022−2⋅0⋅0=0,V(2,0)=−(2)22−2⋅2⋅0=−42=−2.V(0,0)= -\frac{0^2}{2}-2\cdot0\cdot0=0,\quad V(2,0)= -\frac{(2)^2}{2}-2\cdot2\cdot0 = -\frac{4}{2} = -2. ΔVOP=V(2,0)−V(0,0)=−2−0=−2.\Delta V_{OP}=V(2,0)-V(0,0)=-2-0=-2.

    Thus,

    WOP=q ΔVOP=(−10×10−6)(−2)=+20×10−6 J.W_{OP}=q\,\Delta V_{OP}= (-10\times10^{-6})(-2)=+20\times10^{-6}\,J.

  • Path PQPQ: (from P(2,0)P(2,0) to Q(2,2)Q(2,2))

    V(2,0)=−2(from above),V(2,2)=−(2)22−2⋅2⋅2=−42−8=−2−8=−10.V(2,0)=-2 \quad \text{(from above)},\quad V(2,2)= -\frac{(2)^2}{2}-2\cdot2\cdot2 = -\frac{4}{2} -8 = -2 -8=-10. ΔVPQ=V(2,2)−V(2,0)=−10−(−2)=−8.\Delta V_{PQ}= V(2,2)-V(2,0)= -10 - (-2)=-8.

    Then,

    WPQ=q ΔVPQ=(−10×10−6)(−8)=+80×10−6 J.W_{PQ}=q\,\Delta V_{PQ}= (-10\times10^{-6})(-8)=+80\times10^{-6}\,J.

  • Path QOQO: (from Q(2,2)Q(2,2) to O(0,0)O(0,0))

    V(2,2)=−10,V(0,0)=0,V(2,2)=-10,\quad V(0,0)=0,

    so,

    ΔVQO=V(0,0)−V(2,2)=0−(−10)=+10.\Delta V_{QO}=V(0,0)-V(2,2)=0-(-10)=+10.

    Hence,

    WQO=q ΔVQO=(−10×10−6)(10)=−100×10−6 J.W_{QO}= q\,\Delta V_{QO}= (-10\times10^{-6})(10)= -100\times10^{-6}\,J.
  1. Round Trip:

The total work done by the external agent over the closed path OP→PQ→QOOP \rightarrow PQ \rightarrow QO is:

Wtotal=WOP+WPQ+WQO=20×10−6+80×10−6−100×10−6=0 J.W_{\text{total}}=W_{OP}+W_{PQ}+W_{QO}=20\times10^{-6}+80\times10^{-6} -100\times10^{-6}=0\,J.

(This is expected since the field is conservative.)