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Is there a point along the line joining two equal positive charges where the electric field is zero? Where the electric potential is zero? Explain.

Short Answer

Expert verified

Yes, the electric field is zero at the midpoint of the line joining of two equal positive charges.

No, the electric potential along the line joining of two equal positive charges can not be zero.

Step by step solution

01

Understanding the electric field and electric potential due to a point charge

The electric field at any point due to a point charge relies on the magnitude of charge and square of the distance from the charge, whereas the electric potential relies on the magnitude of charge and distance from the charge.

02

Evaluation of the point where the electric field is zero

The expression for the electric field due to a point charge is,

\(\overrightarrow E = k\frac{q}{{{r^2}}}\hat r\)

Here, k is Coulomb’s constant, q is the charge and r is the distance.

The electric field is a vector quantity. Since both the charges are identical, the magnitude of the electric field at the midpoint of the line joining the charges is the same but opposite. So, they will cancel out each other and the net electric field is zero.

Thus, the electric field is zero at the midpoint of the line joining of two equal positive charges.

03

Evaluation of the point where the electric potential is zero

The expression for the electric potential due to a point charge is,

\(V = k\frac{q}{r}\)

The electric potential is a scalar quantity. Since both the charges are positive, therefore the electric potential at any point is the sum of the electric potentials of both the charges, which is a positive quantity.

Thus, the electric potential along the line joining of two equal positive charges can not be zero.

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Most popular questions from this chapter

If a negative charge is initially at rest in an electric field, will it move toward a region of higher potential or lower potential? What about a positive charge? How does the potential energy of the charge change in each instance? Explain.

(II) Three point charges are arranged at the corners of a square of side l as shown in Fig. 17–39. What is the potential at the fourth corner (point A)?

FIGURE 17–39 Problem 22.

(II) An electric field greater than about \({\bf{3 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{V/m}}\)causes air to break down (electrons are removed from the atoms and then recombine, emitting light). See Section 17–2 andTable 17–3. If you shuffle along a carpet and then reach for a doorknob, a spark flies across a gap you estimate to be 1 mm between your finger and the doorknob. Estimate the voltage between your finger and the doorknob. Why is no harm done?

(II) Point a is 62 cm north of a \( - {\bf{3}}{\bf{.8}}\;{\bf{\mu C}}\) point charge, and point b is 88 cm west of the charge (Fig. 17–40). Determine (a) \({{\bf{V}}_{\bf{b}}} - {{\bf{V}}_{\bf{a}}}\) and (b) \({{\bf{\vec E}}_{\bf{b}}} - {{\bf{\vec E}}_{\bf{a}}}\) (magnitude and direction).

FIGURE 17–40 Problem 27.

Four identical point charges are arranged at the corners of a square [Hint: Draw a figure]. The electric field E and potential V at the centre of the square are

(a) \(E = 0\), \(V = 0\).

(b) \(E = 0\), \(V \ne 0\).

(c) \(E \ne 0\), \(V \ne 0\).

(d) \(E \ne 0\), \(V = 0\).

(e) \(E = V\) regardless of the value.

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