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Chapter 26: Q. 7 (page 737)

What are the magnitude and direction of the electric field at the dot in Figure EX26.7?

Short Answer

Expert verified

The magnitude is 20000V/m and direction of the electric field isvertically downwards.

Step by step solution

01

Given information and formula used  

Given figure :

Theory used :

The electric field is directed towards the decreasing potential and has a magnitude equal to the rate of change of the potential.

The electric field will be equal to the potential difference divided by the distance between these potentials. That is to say,

E=∆Vd

02

Determining the magnitude and direction of the electric field at the dot in the figure

The potential varies by200Vover a distance of1cm=0.01min our case, hence the field's magnitude is :

E=200V0.01m=20000V/m

The field points downwards as the potential falls in the downward direction.

So, the Electric field points vertically downwards.

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Most popular questions from this chapter

The electric potential is 40Vat point A near a uniformly charged sphere. At point B, 2.0μmfarther away from the sphere, the potential has decreased by 0.16 mV. How far is point A from the center of the sphere?

What is the capacitance of the two metal spheres shown in Figure EX26.22?

Metal sphere 1 has a positive charge of 6.0 nC. Metal sphere 2, which is twice the diameter of sphere 1, is initially uncharged. The spheres are then connected together by a long, thin metal wire. What are the final charges on each sphere?

An isolated 5.0μF parallel-plate capacitor has 4.0mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.0μF. How much work is done by the external force?

A 6mFcapacitor, a 10mFcapacitor, and a 16mFcapacitor are connected in parallel. What is their equivalent capacitance?

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