Electric KB1 Wrap Sunglasses,Gloss Black Frame/Grey Lens,one size

Electric KB1 Wrap Sunglasses,Gloss Black Frame/Grey Lens,one size

Electric - click on the image below for more information.

• Case included
• Lenses are prescription ready (Rx-able)
• Size category 3
• Kyle Bush signature frame

Electric

Your Electric KB1 sunglasses are a part of who you are, and the thought of facing the bright, over-exposed world without them makes you tremble. You just don't take them off. The KB1 has a scratch-resistant, nearly indestructible construction that can keep up with you no matter how crazy things get, and the ergonomic shape keeps means they will feel great even if you insist on wearing them forever.

Product FeaturesFrame: 8-base grilamidHinge: stainless steelLens: polycarbonateInterchangeable Lens: noPolarized: noFace Size: Case Type: softNose Pads: noTemple Pads: noRecommended Use: streetwearManufacturer Warranty: 1 yearFrame Measurements:

Electric KB1 Wrap Sunglasses,Gloss Black Frame/Grey Lens,one size

Click on the button for more Electric information and reviews.

World's Columbian Exposition: Electric Fountain, Chicago, United States, 1893.


Image by Brooklyn Museum

Music video by MGMT performing Electric Feel. YouTube view counts pre-VEVO: 1337695 (C) 2008 Sony Music Entertainment
Video Rating: 4 / 5

What is the electric potential at the point on the x-axis where the electric field is zero?

Best answer:
it also bcomes 0Electric potential from a charge is K*q/r. If we consider the -10nC charge at the origin, then the potential from it is -K*10/|x|; The potential from the 20nC charge is +K*20/|x-14|. They must sum to zero, so 20/|x-14| - 10/|x| = 0, 20*|x| = 10*|x-14| 2*|x| = |x - 14|

2x = x - 14 or 2x = 14 - x

x = -14 x = 14/3 Plug these in and you'll see they both result in 0 potential at those points.

Only he value of x = 14/3 is between the charges. The field from the -10nC charge is toward the origin, and the field from the 10nC charge is also toward the origin. The fields will then sum directly

The electric field is

E = [10/(0.14/3)^2 + 20/(14/3 - 14)^2]*K*10^-9

K = 9*10^9 N*m^2/C, so

E = [10/(0.14/3)^2 + 20/(0.14/3 - 0.14)^2]*9

E = 62*10^3 N/C

The field points to the -10nC chargeLet q1 = -10.0 nC , and q2 = 20.0 nC
E1(P) + E2(P) = 0
|E1| = |E2|
|q1|/r1^2 = q2/r2^2
r2/r1 = sqr(q2/|q1|) = sqr 2
r2 = r1 + d
Solving : r1 = d/(sqr2 -1) = 33.8 cm ; r2 = 47.8 cm
(a) V = (1/4pi €o)(q2/r2 + q1/r1) = 110 V
(b) V=0 |q1|/r1 = q2/r2
r2 = 2 r1 ; r2 + r1 = d
solving
r2 = 28/3 cm ; r1 = 14/3 cm
E = E1 + E2 = (1/4pi€o)(|q1|/r1^2 + q2/r2^2)
E is // at X axis and direct from q2 to q1.

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