Question

So we have this sphere with a charge of 7.0 nC sitting at the origin of our x-y plane. Another sphere with a charge of -3.0 nC is at a position of 4.00 mm above the origin.

What is the electric potential at the coordinate (36.0 mm, 0.00 mm)?

Now we fire a proton along the x-axis from infinity. By the time it gets to the position (36.0 mm, 0.00 mm), the proton has stopped. What must have been its initial speed?

So we have this sphere with a charge of 7.0 nC sitting at the origin of our x-y plane. Another sphere with a charge of -3.0 nC is at a position of 4.00 mm above the origin. What is the electric potential at the coordinate (36.0 mm, 0.00 mm)? Now we fire a proton along the x-axis from infinity. By the time it gets to the position (36.0 mm, 0.00 mm), the proton has stopped. What must have been its initial speed?

Answer 1

a -3. Onc 10,4) B There is a shhere A with charge the situated at origin (0,0) and there is another sphere with chage - 3:0 nc situated 4 mm above at co-wedinates (0,4). 4.00mm Ad 70nc (36.0 mm, 0) 36.0mm Thelectric potential due to charge q at distance r is defined as, V = 4 1hE. Here to is the permittivity of free space. 36mm The electric potential at point p can be calcuted by taking sum of electric potential at point P due to individul spheres. The electric potential at paint p due to spher A is, V = dato Substitute 9x10 m for the tone for q, and 36.0mm for r,. V , = 9 x 10 / 2100 nC = (2x 109 Nm/2). (1x100 (36x103) V = 1.75X10° V – @ The electric field due to sphere B is, ... V2 = 1 into a Sublitule 9x10 Nmle for 1 - 3. One for q and a forr V₂ = (9x109 m²/2) + (-3 x 103 c). The value of a can be calculated by pythagorus theorem, Rp² = RQ² + Q P2 x = (4 mmp + (36 mm)2 = 1312 mm2 =(36.22mm)2 x = 36.22mm x = 36.22x100m (-3x109 c) Therefore, I ₂ = (3x10 Nm / 2) * 36 22 x 103 m = -0.745 x 103 v

Therefor V, = 1075 x 10 3 v V₂ = -0.145 x 10 3 v The net potential at point p ý V=V, + V₂ = (1.75 -1.745) x103 v =(1.005 X 103) v Iv=1.005 x 103 v Therefor electre potential at co-ordinate (36.0mm, o) is 1.00 5x108 v) Auffase a fraton is moving towards origin from infinity with initial speed &. The proton stopped at point P (36 mm, o). - Initially the proton is at infinity and the potential energy at infornity is zero. So proton only posses kinetic energy at as. Finally proton stopped at point P. Hence it's velocity become zero Do it's Kinetic energy become zero. Therefvee according to energy consewation, all 'Kintic energy converted to Putenticul energy. dcoreling to energy conservation. Einitially = Efinally, Ko Eino = Piffin £m re2 = qov Here mp is the mass of proton (1.6726x1027 kg) and a is the charge on poroton (106710-13). +11.6726x102kg x v² = 0.6x10 19 c) x6.005x103 v) e² - 2x11. 641079 c) x (1.0054 103 v) 1.6726 X1027 kg v² = 1.922 x 10" m/s² v2 = 19.22 4 10 10 m/s² v = $19.22x1010 m/s = 4038 x 10 5 m/s Hence initial speed of proton is ſo = 4 .38 x 105 m/s Hence,