JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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.....\(V\)
- A \(320\)
- B \(-48\)
- C \(-520\)
- D \(180\)
Answer & Solution
Correct Answer
(D) \(180\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{E}}=(20 x+10) \hat{\mathrm{i}}\) \(\mathrm{V}_{1}-\mathrm{V}_{2}=-\int_{-5}^{1}(20 \mathrm{x}+10) \mathrm{d} \mathrm{x}\) \(\mathrm{V}_{1}-\mathrm{V}_{2}=-\left(10 \mathrm{x}^{2}+10 \mathrm{x}\right)_{-5}^{1}\)…
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