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Chapter 11: Problem 1582

The electric potential \(\mathrm{V}\) is given as a function of distance \(\mathrm{x}\) (meter) by $\mathrm{V}=\left(5 \mathrm{x}^{2}+10 \mathrm{x}-9\right)\( volt. Value of electric field at \)\mathrm{x}=1$ is \(\ldots \ldots\) \((\mathrm{A})-20(\mathrm{v} / \mathrm{m})\) (B) \(6(\mathrm{v} / \mathrm{m})\) (C) \(11(\mathrm{v} / \mathrm{m})\) (D) \(-23(\mathrm{v} / \mathrm{m})\)

Short Answer

Expert verified
The electric field E at x = 1 is -20 V/m. The correct answer is (A) -20 V/m.

Step by step solution

01

Find the derivative dV/dx of the potential V(x)

To find the derivative of V(x) with respect to x, differentiate the function with respect to x: \(V(x) = 5x^2 + 10x - 9\) Taking the derivative, we get: \(E = -\frac{dV}{dx} = -\frac{d(5x^2 + 10x - 9)}{dx}\)
02

Simplify the expression of the electric field E

Now we need to simplify the expression for E: Applying differentiation rules: \(E = - (10x + 10)\) Now we have the electric field E as a function of distance x.
03

Evaluate the electric field E at x = 1

We are asked to find the value of the electric field at x = 1. To do this, substitute the value x = 1 in the equation we found in Step 2: \(E(1) = -(10(1) + 10) = -20\) The value of the electric field E at x = 1 is -20 V/m. The correct answer is (A) -20 V/m.

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Most popular questions from this chapter

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)

A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged $2.5 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)
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