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jenyasd209 [6]
10 months ago
15

A +13.4 nC charge is located at (0,9.4) cm and a -4.23 nC charge is located (4.99, 0) cm. Where would a -14.23 nC charge need to

be located in order that the electric field at the origin be zero? Express your answer, in cm, as the magnitude of the distance of q3 from the origin.
Physics
1 answer:
antiseptic1488 [7]10 months ago
4 0

In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge

E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis

E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis

The red arrow shows the direction of which the electric field points.

To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2

Etotal = sqrt(E1+E2) = 20494.97 V/m

E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2

d = 0.079 m = 7.9 cm

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3 years ago
A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 790.5 meters from the wall and the sound of h
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4.80 seconds

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A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
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Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

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2 years ago
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julsineya [31]

Explanation:

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Using dimensional of A, B and C in above formula. So,

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Comparing the powers both sides,

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Now, solving equation (1) and (2) we get :

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Hence, the correct option is (E).

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