3 years ago

I WILL GIVE BRAINLIEST IF ANSWER FAST What is the measurement on this Dial Caliper?

Engineering

1 answer:

garik1379 [7]3 years ago

8 0

Answer:

b i think i dont see any dial caliper

Explanation:

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Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg

babunello [35]

Answer:

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

$\sigma_A = -4.10^{-5}C/m^2$

$\sigma_B= -7.10^{-5}C/m^2$

$\sigma_C = -3.1^{-5}C/m^2$

We find the electric field for each one, this formula is given by,

$E= \frac{\sigma_i}{2\epsilon_0}$

Substituting each value from the three charged sheets, we have

$E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})$

$E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

The electric field is

$E_{NET}= E_A+E_B+E_C$

$E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_{NET} = 0$

Force on each sheet is,

$F=E_{NET}\sigma ds$

$F=0$

The total force is 0

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3 years ago