An electricity and magnetism problem by A Former Brilliant Member

Electricity and Magnetism Level 3

Find the potential difference V A B V_{AB} between A ( 0 , 0 , 0 ) A(0,0,0) and B ( 1 , 2 , 3 ) B(1,2,3) in an Electric Field

( i ) E → (i) E^{\rightarrow } = ( y i + x j ) V m − 1 = (yi + xj)Vm^{-1}

( i i ) E → = ( 3 x 2 y i + x 3 j ) V m − 1 (ii) E^{\rightarrow } = (3x^2y i + x^3 j) Vm^{-1}

Where i i and j j are unit vectors along x and y - axes respectively.

Find the sum of potentials in case ( i ) (i) and ( i i ) (ii)


The answer is 4.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

We know that d V = − E • d r dV = -E•dr

Here, d r → = d x i + d y j + d z k dr^{\rightarrow } = dx i + dy j + dz k

( i ) d V = − ( y i + x j ) • ( d x i + d y j + d z k ) = − ( y d x + x d y ) = − d ( x y ) (i) dV = -(y i + xj)•( dxi + dy j + dz k) = -(ydx + xdy) = -d(xy)

V A B = − ∫ ( 1 , 2 , 3 ) ( 0 , 0 , 0 ) d ( x y ) = [ x y ] ( 1 , 2 , 3 ) ( 0 , 0 , 0 ) = 2 V V_{AB} = -\displaystyle \int_{(1,2,3)}^{(0,0,0)} d(xy) = \begin{bmatrix} xy\end{bmatrix}_{(1,2,3)}^{(0,0,0)} = 2V

( i i ) d V = − ( 3 x 2 y i + x 3 j ) • ( d x i + d y j + d z k ) = − ( 3 x 2 y + x 3 d y ) = − d ( x 3 y ) (ii) dV = -(3x^2y i +x^3 j)•(dx i + dy j + dz k) = -(3x^2y + x^3dy)= -d(x^3y)

V A B = − ∫ ( 1.2.3 ) ( 0 , 0 , 0 , ) d ( x 3 y ) = − [ ( x 3 y ) ] ( 1.2.3 ) ( 0 , 0 , 0 , ) = 2 V V_{AB} = -\displaystyle \int_{(1.2.3)}^{(0,0,0,)} d(x^3y) = - \begin{bmatrix}(x^3y)\end{bmatrix}_{(1.2.3)}^{(0,0,0,)} = 2V

Hence, A N S W E R = 2 + 2 = 4 V ANSWER = 2 + 2= \boxed {4V}

2 Helpful 0 Interesting 0 Brilliant 1 Confused

The question seemed to have a decimal between the 1 and 2 for point B. The answer I got was 8.784, because I thought it was just giving the x and y coordinates lol.

Tristan Goodman - 2 years, 4 months ago

Thanks for pointing that out, I have corrected that.

A Former Brilliant Member - 2 years, 4 months ago

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You're welcome.

Tristan Goodman - 2 years, 4 months ago

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