Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Question: Figure 24-47 shows a thin plastic rod of length L = 12.0 cmand uniform positive charge Q = 56.1fClying on an xaxis. With V = 0at infinity, find the electric potential at point P1 on the axis, at distance d = 250 cmfrom the rod.

Short Answer

Expert verified

Answer:

The electric potential at point P1 on the axis is 7.396×10-3V.

Step by step solution

01

The given data

  1. Length of the rod, L = 12cm
  2. Uniform positive charge on the axis, Q = 56.1fC
  3. The electric potential at V = 0 is infinity.
  4. The distance of the point, d = 2.50 cm
02

Understanding the concept of the electric potential

Substituting the value of charge from the concept of the linear charge density in the formula of the electric potential due to a charge at a point, we can get the required value of the potential on the x-axis by using the given data.

Formulae:

The linear charge density of a distribution, λ=dqdx (i)

The electric potential at a point due to a point charge, dV=14πε0dqx+d (ii)

03

Calculation of the electric potential

Consider an infinitesimal segment dx that has a charge.

The expression for the electric potential due to the small element of the finite line of charge along the direction of the line charge is given using the equation (i) and then the given values in equation (ii) as follows:

dV=14πε0λdxx+d

Now, the electric potential is given by integrating the above value and substituting the given values as follows:(Here, the length of the charge is changing from 0 to L)

Therefore, the value of the potential at point P is 7.396 x 10-3V

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A plastic rod has been bent into a circle of radius R = 8.20 cm. It has a charge Q1 = +4.20pCuniformly distributed along one-quarter of its circumference and a charge Q2 = -6Q1uniformly distributed along the rest of the circumference (Fig. 24-44). With V = 0at infinity, what is the electric potential at (a) the center Cof the circle and (b) point P, on the central axis of the circle at distance D = 6.71cmfrom the center?

A thick spherical shell of charge Q and uniform volume charge density r is bounded by radiir1and r2>r1.With V=0 at infinity, find the electric potential V as a function of distance r from the center of the distribution, considering regions

(a)r>r2 ,

(b)r2>r>r1 , and

(c)r<r1 .

(d) Do these solutions agree with each other at r=r2andr=r1? (Hint: See Module 23-6.)

A spherical drop of water carrying a charge of 30 pChas a potential of 500 Vat its surface (with V = 0at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

Two uniformly charged, infinite, nonconducting planes are parallel to a yz plane and positioned at x = -50cmand x =+ 50cm. The charge densities on the planes are -50 nC/m2and +25 nC/m2 , respectively. What is the magnitude of the potential difference between the origin and the point on the x axis at x = +80cm ?

An electric field of approximately 100 V/mis often observed near the surface of Earth. If this were the field over the entire surface, what would be the electric potential of a point on the surface? (SetV=0 at infinity.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free