 diferira7c

2022-01-07

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. (a) If the surface charge density for each plate has magnitude $47.0\mathrm{nC}/{\mathrm{m}}^{2}$, what is the magnitude of $\stackrel{\to }{E}$ in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference? Fasaniu

Step 1
Given:
${\sigma }_{a}=47.0nC\cdot {m}^{2}$
${\sigma }_{b}=-47.0nC\cdot {m}^{2}$
d=2.2 cm
Required
a)E betwween them.
b)V between them.
c)If d becomes 2d find E and V.
Step 2
a)According to electric field for parallel plstes the electric field of infinite parallel plates is given by
$E=\frac{\sigma }{2{\in }_{0}}⇒\left(1\right)$
So the field is independent on d so the net field is
$E={E}_{1}+{E}_{2}=\frac{\sigma }{2{\in }_{0}}+\frac{\sigma }{2{\in }_{0}}=\begin{array}{|c|}\hline \frac{\sigma }{{\in }_{0}}\\ \hline\end{array}$
Substitution in (1) yields
$E=\frac{47×{10}^{-9}}{8.85×{10}^{-12}}=\begin{array}{|c|}\hline 5.3×{10}^{3}N/C\\ \hline\end{array}$ Elois Puryear

Step 3
b)According to example (23.9) Since the electric potential between tham at any point and is given by
$V=Ed⇒\left(2\right)$
So Sumstitution in (2) yields
$V=2.2×{10}^{-2}×5.3×{10}^{3}=\begin{array}{|c|}\hline 116.6V\\ \hline\end{array}$ karton

Step 4
c)If the distance between plantes is doubled it wiil make no changes on the field because the field is independent on d.For the voltage, according to (2) the voltage depends on d so if d is doubled then the potential between plantes must be doubled.
Result
a) $E=5.3×{10}^{3}N/C$
b) $V=116.6V$
c) E is independent on d so it won`t change, but the potential depends on d so if d is doubled then V must be doubled.

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