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bazaltina [42]
3 years ago
9

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

zero? Express your answer in terms of some or all of the variables s, q1, q2 and k =14πϵ0. If your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work.
Physics
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

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A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug
vlada-n [284]

Answer:

The Minimum wavelength is  \lambda_{min}= 382.2nm

The Maximum wavelength is \lambda_{max}= 624.2nm

Explanation:

From the question we are told that  

              The energy range is  E_r = 3.25eV \ and  \ 19.9eV

   Considering E = 19.9eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 19.9 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{max}}

Here h is the Planck's constant with value of  h= 6.625*10^{-34}J\cdot s

        c is the speed of light with value of  c = 3*10^8 m/s

Substituting values and making \lambda the subject of the formula

                       \lambda_{max} = \frac{hc}{E}

                         = \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}

                         \lambda_{max}= 624.2nm

  Considering E = 3.25eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 3.25 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{min}}

Substituting values and making \lambda the subject of the formula

                       \lambda_{min} = \frac{hc}{E}

                           = \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}

                           \lambda_{min}= 382.2nm

3 0
3 years ago
Your friend has offered you a ride home from school. During the ride home, you realize that your friend is clearly distracted an
inn [45]
Step 1: Tell them to keep their eyes on the road.Step 2: Watch them to make sure they keep their eyes on the road.Step 3: Don't converse with them because you might risk distracting them.
It makes an awkward situation but it ensures that they'll feel so pressured to keep their eyes on the road or pull over so you can call a different ride.
6 0
3 years ago
Read 2 more answers
What will be the displacement of this spring if the load is 4n with a constant of 5n\m
nikdorinn [45]

As we have formula for mass spring system

F= kx           ( where k is spring constant , x is displacement , F is the force acting due to load)

To find =x

So  x=F/k

x= (4N) / (5N/m)

x= 0.8 m

3 0
3 years ago
In a stormy sea, 2 waves pass a fixed point every second, and the waves are 10 meters apart. What is the speed of the wave?
Kitty [74]

             Speed of a wave  =  (frequency) x (wavelength)

                                         =  (2 per second) x (10 m)

                                         =      20 m/s .

7 0
4 years ago
20
fomenos

Answer:

The correct option is;

(A) 1 m/s

Explanation:

Given that Ramu and Somu are running towards North with 3 m/s and 4 m/s respectively.

Their friend Srinu is running towards South with a speed of 2 m/s

The magnitude of the relative velocity of Ramu with respect to Ramu is found as follows;

Velocity of Ramu = 3 m/s North ↑

Velocity of Srinu = 2 m/s South↓ = -2 m/s North ↑

Relative velocity of Ramu to Srinu = Velocity of Ramu North + Velocity of Srinu North

Relative velocity of Ramu to Srinu = 3 m/s + (-2) m/s = (3 - 2) m/s = 1 m/s

Relative velocity of Ramu to Srinu = 1 m/s.

4 0
4 years ago
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