NettetThe alternative way is looking at the original triangle and observing that the horizontal component is 3/5 of the hypotenuse. So the electric field in the horizontal direction is 3/5 of the original electric field (2.88 N/C in our case). (3/5)x (2.88 N/C)=1.728 N/C Dave rounded that off to 1.73 N/C 1 comment ( 22 votes) Upvote Downvote Flag more NettetClick here👆to get an answer to your question ️ A charged particle of mass 1 kg and charge 2 C is thrown from a horizontal ground at an angle 0 = 45° with speed 20 m/s. In space a horizontal clectric field E = 2 x 107 v/m exist. The range on ma horizontal ground of the projectile thrown is .... X (1) 100 m (2) 50 m (3) 200 m (4) 0 m he charge +

5.14: Electric Field as the Gradient of Potential

Nettet12. sep. 2024 · The electric field for a surface charge is given by →E(P) = 1 4πϵ0∫surfaceσdA r2 ˆr. To solve surface charge problems, we break the surface into … Nettet#physics #class12 #youtube #youtubevideo Through this lecture we are going to discuss and understand about Electric Field and the Electric Field Intensity ... body electric workshops

A charged particle of mass m = 2 kg and charge q = 2 mu C is

NettetA spherical conductor of radius 10 cm has a charge of 3.2×10 −7C distributed uniformly. What is the magnitude of electric field at a point 15cm from the centre of the sphere? … Nettet12. sep. 2024 · where we have used the fact that 1 N·m = 1 joule (J) of energy and 1 J/C = 1 V. Electric field intensity is a Vector Field Electric field intensity ( E, N/C or V/m) … Nettet22. nov. 2024 · Solution: the electric potential difference \Delta V ΔV between two points where a uniform electric field E E exists is related together by E=\frac {\Delta V} {d} E = dΔV where d d is the distance between those points. Substituting the numerical values, we will have E=\frac {240} {2.4}=100\,\rm V/m E = 2.4240 = 100V/m Note that the volt per ... glazed corned beef recipes oven