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Vlad [161]
3 years ago
15

Identical +7.67 μC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed

to one of the empty corners, so that the total potential at the remaining empty corner is 0 V?
Physics
1 answer:
Svetlanka [38]3 years ago
8 0

Answer:

q₃ = - 13.0935 μC

Explanation:

Given

q₁ = q₂ = +7.67 μC

We use the equation

V = Kq/r

We can apply it as follows

V₁ = K*q₁/r₁ = K*q₁/(√2*L)

V₂ = K*q₂/r₂ = K*q₂/L

V₃ = K*q₃/r₃ = K*q₃/L

Then

V₁ + V₂ + V₃ = 0

⇒   (K*q₁/(√2*L)) + (K*q₂/L) + (K*q₃/L) = 0

⇒   (K/L)*((q₁/√2) + q₂ + q₃) = 0

⇒   (q₁/√2) + q₂ + q₃ = 0

Since q₁ = q₂

⇒   (q₁)((1/√2) + 1) + q₃ = 0

⇒   q₃ = - (q₁)((1/√2) + 1) = +7.67 μC*(1.7071)

⇒   q₃ = - 13.0935 μC

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What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh
Goshia [24]

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

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3 0
3 years ago
A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.
UkoKoshka [18]

Answer:

a) 1.6 mN  b) -1.6 mN  c) -1.6 mN  d) 1.6 mN

Explanation:

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a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ =  1.6 mN (going in the positive x direction, towards q₂)

6 0
3 years ago
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