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liraira [26]
2 years ago
12

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

ial difference would appear as:_________
Physics
1 answer:
vfiekz [6]2 years ago
5 0

The potential difference comes out to be

10 \times 10 {}^{ - 3} m

Given:

σ = 8. 85 × 10-9 c/m2

we know,

E = \frac{σ}{2ε0}

E =  \frac{8.85 \times 10 {}^{ - 9} }{2ε0}

E =  \frac{v}{d}

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

=  \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }

Δ = 10 \times 10 {}^{ - 3} m

Thus the potential difference is

10 \times 10 {}^{ - 3} m

Learn more about potential difference from here: brainly.com/question/28165869

#SPJ4

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7 0
3 years ago
The valu
natulia [17]

Answer:

\% Error = 2.6\%

Explanation:

Given

x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45

Required

Determine the percentage error

First, we calculate the mean

\bar x = \frac{\sum x}{n}

This gives:

\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}

\bar x = \frac{9.06}{6}

\bar x = 1.51

Next, calculate the mean absolute error (E)

|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}

This gives:

|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}

|E| = \sqrt{\frac{1}{6}*0.0132}

|E| = \sqrt{0.0022}

|E| = 0.04

Next, calculate the relative error (R)

R = \frac{|E|}{\bar x}

R = \frac{0.04}{1.51}

R = 0.026

Lastly, the percentage error is calculated as:

\% Error = R * 100\%

\% Error = 0.026 * 100\%

\% Error = 2.6\%

4 0
3 years ago
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