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dalvyx [7]
2 years ago
11

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

R away from the wire using Gauss's Law for Electrostatics.
Physics
1 answer:
marshall27 [118]2 years ago
4 0

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

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2 years ago
A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

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2 years ago
A box is at rest on an inclined surface. Which diagram correctly illustrates all forces, including any component forces?
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Answer:

The answer is D.

Explanation:

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g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
2 years ago
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