विधुत क्षेत्र रेखाएं Electric Field Lines | वैधुत आवेश तथा क्षेत्र Class 12 Ch 1 Lec 4 | Gopal Sir

Class 12 Physics (Hindi)| Chapter 1 वैद्युत आवेश तथा क्षेत्र | Important Questions | UP Boards

Electric Due To Point Charge (बिन्दु आवेश के कारण क्षेत्र)|Electric Potential (वेधयुत विभव)|Questions (प्रश्न)|Electric Potential Due To Hollow Charged Sphere (खोखले आवेशित गोले के कारण विभव)|Potential Energy Between Two Point Charges (दो बिन्दु आवेशों के मध्य संचित ऊर्जा)|Question (प्रश्न)|Dipole (द्विध्रुव)|Potential Difference In Uniform Electric Field (एक समान क्षेत्र मे विभावांतर)|OMR

Two point charges 4Q, Q are separated by 1m in air. At what point on the line joining the charges is the electric field intensity zero ? Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q=2xx10^(-7)C .

Point charges – q, 2q, – 3q, q, – q, 2q, – 3q, q, – q, 2q, – 3q , and q have been placed at marks 1, 2, 3, 4, 5 .....12 respectively on the circular dial of a clock. Find the electric field intensity at the centre of the dial if distance of each charge from the centre is r.

बिंदु आवेश के कारण वैद्युत क्षेत्र की तीरवता|वैद्युत बल रेखाएँ|एकसमान तथा असमान वैद्युत क्षेत्र में वैद्युत बल रेखाएं|OMR

In moving from A to B along an electric field line, the work done by the electric field on an electron is 6.4 xx 10^-19 J . If phi_1 and phi_2 are equipotential surfaces, then the potential difference V_C - V_A is. .

A large charged metal sheet is placed in a uniform electric field, perpendicular to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is E_1 = 5 xx 10^5 Vm^(-1) and on the right it is E_2 = 3 xx 10^5 V m^(-1) . The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet. Use (1/(4piepsilon_0)) = 9 xx 10^(9) Nm^(2)C^(-2)

(A) Two electric field lines cannot cross each other. Also, they cannot crossed loops. Give reasons. (B) A particle of charge 2 muC and mass 1.6 g is moving with a velocity 4 hati ms^(-1) .At t=0 the particle enters in a region having an electric field vecE ( in NC^(-1) ) = 80hati + 60hatj . Find the velocity of the particle at t =5 s

Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2 xx 10^(-6) C m^(-1) .