4 the electric field at 2 cm from a long straight wire is 20...

4. The electric field at 2 cm from a long straight wire is 20 N/C. The electric field at 5 cm from the wire is a) 4 N/C b) 2 N/C c) 0.125 N/C d) 8 N/C $$ \frac{E_1}{E_2} = \frac{r_2}{r_1} \Rightarrow \frac{20}{E_1} = \frac{5}{2} $$

Transcript

00:01 Hello students in this question what is given what is given is that we have a long straight conductor with the uniform magnet electric field right electric field uh electric field uniform charge distribution lambda is there so we are given at two centimeter from the surface the electric field is 20 newton per cool ome okay so that is two centimeters from the surface the electric field is 20 newton per cool ome okay so that is two centimeters and at 4 centimeters what is the electric field we need to find out that's the first question so let's solve that so e can be written down as e is equal to lambda by 2 pi epsilon 0 r for uniformly distributed long straight conductor where lambda is the charge distribution and a epsilon zero is the permittivity and r is the distance from the from the conductor so what we can do now is that we can substitute the values substitute the values here so we know it is 20 newton per kool -oam must be equal to lambda by 2 pi epsilon 0 0 meters right 2 centimeters so now lambda will be equal to 20 times 2 pi times epsilon 0 times 0 .02 okay so that's it so what it means is that we can calculate lambda now lambda will be equal to 20 times 2 pi times epsilon is 8 .854 times 10 to the power minus 12 times 0 .02 so that will give you a value of 2 .25 times 10 to the power minus 11 koum per meter okay that is lambda so we found out lambda now let's find out for for r is equal to 0 .04 meter or 4 centimeter what is the electric field so electric field is equal to lambda which is 2 .25 times 10 to the power minus 11 divided by 2 by 2 by epsilon 0 is just 8 .54 times okay 10 to the bar minus 12 times r is 0 .04 now computing that you'll get e is equal to 10 newton per kolo so that's the solution for first question e is equal to 10 newton per kulum at 4 centimeters now going to the second question we have a long infinite long charge distribution like this okay and from two centimeters from the charge right it is given that it's two centimeters same but different values it is 3 30 newton per cool so at 4 centimeters what it is okay what it is electric field okay four centimeters what it is so for this the same format but different equation for a uniformly charged surface like this infinitely charged surface like this sigma is a charge density okay the sigma is the charge density so let's write down the equation of electric field e is equal to sigma by 2 pi a 2 epsilon 0 r so that will be the equation now let's use the equation to find out sigma so sigma will be 2 epsilon 0 r times e now substituting the values now we know at 2 centimeters so it is 2 times 8 .854 times 10 to the power minus 12 times 30 times distance is 0 .02 times 30 newton per kulum so that will give you a sigma of 1 .06 times 10 to the power minus 11 koum per meter meter square that's the surface charge distribution now using the sigma we can calculate e so e at r is equal to 0 .4 meter e will be equal to what e will be equal to sigma by 2 epsilon 0 r so that will be sigma is 1 .06 times 10 to the power minus 11 divided by 2 times 8 .854 times 10 to the power minus 12 times 0 .04.

05:49 Okay.

05:50 So computing that you get electric field is equal to 22 .5 newton.

05:59 Oh sorry, newton per cool...

Question

4. The electric field at 2 cm from a long straight wire is 20 N/C. The electric field at 5 cm from the wire is a) 4 N/C b) 2 N/C c) 0.125 N/C d) 8 N/C $$ \frac{E_1}{E_2} = \frac{r_2}{r_1} \Rightarrow \frac{20}{E_1} = \frac{5}{2} $$

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