Question

The magnitude of electric field (in NC-1) in a region varies with the
distance r(in m) as
E = 10 r + 5
By how much does the electric potential increase in moving from
point at r= 1 m to a point at r = 10 m.​

Answer 1

answer : 540 volts

we know, rate of change of electric potential with respect to position is known as electric field at that region.

i.e., E = dV/dr

⇒dV = ∫E dr

⇒∫dV = ∫(10r + 5)dr

⇒V = [5r² + 5r]

putting limits r = 1 to r = 10m

so, V = 5(10)² + 5(10) - (5.1² + 5.1)

= 500 + 50 - 10

= 540 Volts

hence electric potential increases 540 volts in moving from r = 1m to a point at r = 10m.

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Answer 2

$\huge\star\mathbb\blue{{Answer:-}}$

we know, rate of change of electric potential with respect to position is known as electric field at that region.

i.e., E = dV/dr

⇒dV = ∫E dr

⇒∫dV = ∫(10r + 5)dr

⇒V = [5r² + 5r]

putting limits r = 1 to r = 10m

so, V = 5(10)² + 5(10) - (5.1² + 5.1)

= 500 + 50 - 10

= 540 Volts

hence electric potential increases 540 volts in moving from r = 1m to a point at r = 10m.