Verified Solution • 2025–26 Session moderate

QJEEM19Ph2443: JEE Main 2019 Physics - Class 12 JEEM19Ph2443 Solution: The electric field in a region is given by , where E is in NC…

The electric field in a region is given by E→=(Ax+B)i∧\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedge , 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 :-

Multiple Choice Options

  • A –520 V
  • B 180 V
  • C –48 V
  • D 320 V

Step-by-Step Verified Solution

E→=(20x+10)i^\overrightarrow E = (20x + 10)\widehat i

V1−V2=−∫−51(20x+10)dx{V_1} - {V_2} = - \int\limits_{ - 5}^1 {\left( {20x + 10} \right)dx}

V1−V2=−(10x2+10x)−51{V_1} - {V_2} = - \left( {10{x^2} + 10x} \right)_{ - 5}^1

V1−V2=10(25−5−1−1){V_1} - {V_2} = 10\left( {25 - 5 - 1 - 1} \right)

V1−V2=180 V{V_1} - {V_2} = 180\,V

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