2. Three charges equal to -Q, +Q and +Q are located a...

Name: 2. Three charges equal to -Q, +Q and +Q are located a distance a apart along the x axis (see figure 1). The point P is located on the positive y-axis a distance a from the origin. (a) What is the electric field E at point P? (10)
Uploaded: 2023-03-19T21:26:01-08:00
Duration: 6 min 43 s
Channel: Madhur L
Description: 2. Three charges equal to -Q, +Q and +Q are located a distance a apart along the x axis (see figure 1). The point P is located on the positive y-axis a distance a from the origin. (a) What is the electric field E at point P? (10)

2. Three charges equal to -Q, +Q and +Q are located a distance a apart along the x axis (see figure 1). The point P is located on the positive y-axis a distance a from the origin. (a) What is the electric field E at point P? (10)

Transcript

00:01 Hello students in the question we are given three charges electric field due to negative charge it acts towards the charge we have electric field due to positive charges it is away from the charges so here electric field due to this charge it would be in this direction an electric field with this charge would be in this direction we would get net electric field at point where it is equipped to e r p plus e s p plus e tp plus e tp.

00:41 Electric field it is given by formula k constant into charge q divided by r square here r is distance we can write e r p r p r p it is equal to k times minus q divided by distance square that is 2a square dot r cap r cap it is unit vector here we have this equal to k q upon 2a square minus r cap so we have minus r cap equal to minus r cap this is equal to minus a .i.

01:32 Cap plus a .j.

01:34 Cap divided by magnitude that is under root 2a.

01:40 This comes out to be equal to minus i cap upon under root 2 minus j cap upon under root 2.

01:48 So we can write similarly we have esp it is equal to k times charge q divided by distance square that is a square r cap here r cap it is equal to radius vector r divided by magnitude is equal to a j cap divided by a that is j cap then we can write p s p it is equal to k times q divided by a square j cap we have p p p p p p p p p p it is equal to k times charge q divided by distance square that is 2a square r cap here r cap it is equal to radius vector r divided by magnitude this is equal to minus a i cap plus a j cap divided by magnitude that is under root 2a this is out to be equal to minus i cap upon root 2 plus j cap upon root 2.

03:11 So we get etp it is equal to k times q upon 2a square into unit vector that is minus i cap upon root 2 plus j cap upon root 2.

03:27 So finally we can write p p that is sum of these 3 is equal to k times q upon 2a square minus i cap upon root 2 minus j cap on root 2 plus we have k times q upon a square j cap plus we have k times q upon 2 a square 2a square minus i cap upon root 2 plus j cap on upon root 2.

04:10 Sum of 3 vectors we get this equal to electric field at p it is equal to we can add i cap and j cap components so we get here we can also multiply this by 2 and divide this by 2 so we can write a cube upon 2a square common and we have minus i cap upon root 2 minus j cap upon root 2 plus 2 j cap minus i cap upon root 2 plus j cap upon root 2 this comes out to be equal to we have e p over to k q upon 2 a square here we can cancel this we get minus i cap upon root 2 minus i cap upon root 2 plus 2 j cap we have ep, it is equal to kq upon 2a square minus i cap plus 2j cap.

05:23 Here, this is not unit vector.

05:26 So we have magnitude of this vector equal to under root minus 1 square plus 2 square...

Question

2. Three charges equal to -Q, +Q and +Q are located a distance a apart along the x axis (see figure 1). The point P is located on the positive y-axis a distance a from the origin. (a) What is the electric field E at point P? (10)

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