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Sergeu [11.5K]
3 years ago
9

Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t

he values α = 600 N/C and β = 5.00 N⋅m2/C. Points a and b are on the y-axis. .
Point a is at y = 2.00 cm and point b is at y = 3.00 cm. What is the potential difference Va−Vb between these two points.?
Physics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

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