40. ** BIO Electric field of a fish An African fish called...

40. ** BIO Electric field of a fish An African fish called the aba has a charge q = + 1.0 × 10?? C at its head and an equal magnitude negative charge -q at its tail (see Figure P18.40). Determine the magnitude and direction of the electric field at position A and the force exerted on a hydroxide ion (charge -e) at that point. The fish and ion are in water. Indicate any assumptions that you made. Figure P18.40 41. BIO Body cell membrane electric field (a) Determine the average magnitude of the E field across a body cell membrane. A 0.07-V potential difference exists from one side to the other and the membrane is 7.5 × 10?? m thick. (b) Determine the magnitude of the electrical force on a sodium ion (charge +e) in the membrane. Assume that the dielectric constant is 1.0 (it is actually somewhat larger).

Transcript

00:01 Hello, the question is taken from physics and the question is it is taken basically from electricity part.

00:09 An african fish called the abba has a charge of, so african fish has a charge of 10 to the power minus 7 column.

00:18 Okay, at its head an equal magnitude of negative charge at its tail.

00:24 Determine the magnitude and the direction of electric field at point a and the force, inserted on a hydroxide ion e charges minus e at that point.

00:39 The fish and ion are in water indicate any assumption that you made.

00:47 So let us evaluate.

00:52 So this is negative charges here and positive charges here.

00:55 Magnitude of negative and positive charge are already given to us.

00:59 And the distance between these two is 0 .16 meters.

01:03 First we need to evaluate the electric field at this point let us take this is p and it is at a distance 0 .12 meter okay so these are two charges electric field due to positive charge at this point is this direction and due to negative charge the electric field at this point is in the downward direction okay so we can resolve this negative component into two part one is along negative x direction and other is along negative y direction okay so let me evaluate the electric field so this is e1 this is e2 magnitude we already know okay so e1 is equal to the charge is q divided by 4 by epsilon o 10 r square and direction is minus of j the value of q is 10 to the power minus 7 into 9 into 10 to the power 9 divided by r is 0 .12 square into minus of j okay so this will give you value 900 it will be 900 divided by 0 .1 2 square so that will be minus 6250 j0 j0 e2 will be q divided by 4 pi epsilon root r a square let me write this as and direction we need to evaluate we will evaluate the component along x and y direction.

02:50 Let us take this come this is as theta.

02:53 Then this is also theta alternate angles are equal.

02:56 So the component is this is e2 course of theta but in the negative x direction and other component is also in the direction so this is positive okay plus this component is negative the component of v2 okay so that is e2 sine of theta this will be 10 to the power minus 7 into 9 into 10 to the power 9 r is square 2 2 square plus is it in centimeter or meter it is in meter plus 0 .16 square so this will give you the value of e2 is and taking the square of it 0 .12 square plus 0 .16 square taking the square root of it and a reciprocal of it multiply 900 so that is 4 ,500.

04:02 One second whether i did anything wrong with the previous part once again...

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