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Electrostatics question

2019 · 8 Apr · Shift 2 · Q64
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Electrostatics question

2019 · 8 Apr · Shift 2 · Q64

JEE MainPhysicsElectrostaticsMCQ+4 / −1
The electric field in a region is given by E→=(Ax+B)i∧\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedgeE→​=(Ax+B)i∧​ , where E is in NC–1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = –5 is V2, then V1 – V2 is :-
  1. A
    –520 V
  2. B
    180 V
  3. C
    –48 V
  4. D
    320 V
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