E-Field Work

Electricity and Magnetism Level 3

In the x y z xyz coordinate system, an electric field is defined as follows:

E ⃗ = ( E x , E y , E z ) = ( y , z , x ) \large{\vec{E} = (E_x, E_y, E_z) = (y,z,x)}

Suppose one unit of charge moves along a straight-line path from ( 0 , 0 , 0 ) (0,0,0) to ( 1 2 , 1 2 , 0 ) \Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0 \Big) . Determine the work done by the field on the charge. Give your answer as a positive number.


The answer is 0.25.

Steven Chase by Steven Chase , 37, USA

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1 solution

Aaghaz Mahajan
Aug 5, 2019

It is a simple matter of calculating a line integral.

Let r ( t ) = < t 2 , t 2 , 0 > \displaystyle r\left(t\right)=<\frac{t}{\sqrt{2}},\frac{t}{\sqrt{2}},0> denote the straight line path.

Now, because the charge is one unit, the work simply comes out to be

∫ 0 1 E ( r ( t ) ) ⋅ r ′ ( t ) d t \int_0^1E\left(r\left(t\right)\right)\cdot r'\left(t\right)dt

= ∫ 0 1 < t 2 , 0 , t 2 > ⋅ < 1 2 , 1 2 , 0 > d t =\int_0^1<\frac{t}{\sqrt{2}},0,\frac{t}{\sqrt{2}}>\cdot<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0>dt

= ∫ 0 1 t 2 d t =\int_0^1\frac{t}{2}dt

= 1 4 =\frac{1}{4}

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