Physics, 11.03.2020 21:40 mariahrmunoz01

Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.

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Answer from: jwbri

Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the magnitude of the electric field between the plates from the charge density.

b) Determine the potential difference between the plates.

The Electric Field between the two plate is E = 5.536*10^3 N/C

The potential difference between the plate is V = $609 \ Volts$

Explanation:

From the we are given that

The separation between the plate is d = 11.0 cm

The surface charge density is $\sigma = 49.0 n C/m^2 = 49.0*10^{-9}$

Generally Electric field between the plate is mathematically given as

$E = \frac{\sigma }{\epsilon_0}$

Note that $\epsilon_0$ is the permitivity of free space and its value is $8.85 *10 ^{-12} C^2 /N \cdot m^2$

Now substituting values we have

$E = \frac{49*10^{-9}}{8.85*10^{-12}}$

5.537*10^3 N/C

Generally Potential difference between the plate is mathematically given as

V =Ed

Where E is the electric field which is E = 5.536*10^3 N/C

Substituting value we have

V = $= (5.537*10^3 N/C) (11.0*10^{-2}) =609 V$

Answer from: Quest

no. dot product =0 when vecs at right angles

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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