ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2019
  • Medium
  • Numerical

The election field in a region is given by \(\vec E = (Ax + B)\hat i\) where \(E\) is in \(N\,C^{-1}\) and \(x\) in meters. The values of constants are \(A = 20\, SI\, unit\) and  \(B = 10\, SI\, unit\). If the potential at \(x =1\) is \(V_1\) and that at \(x = -5\) is \(V_2\) then \(V_1 -V_2\) is.....\(V\)

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(D) \(180\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{\mathrm{E}}=(20 x+10) \hat{\mathrm{i}}\)

\(\mathrm{V}_{1}-\mathrm{V}_{2}=-\int_{-5}^{1}(20 \mathrm{x}+10) \mathrm{d} \mathrm{x}\)

\(\mathrm{V}_{1}-\mathrm{V}_{2}=-\left(10 \mathrm{x}^{2}+10 \mathrm{x}\right)_{-5}^{1}\)

\(\mathrm{V}_{1}-\mathrm{V}_{2}=10(25-5-1-1)\)

\(V_{1}-V_{2}=180\, V\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app