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Anettt [7]
3 years ago
11

Overview In C, a string is simply an array of characters. We have used literal strings all along – the stuff in between the quot

es in a printf is a literal string. A literal string can be stored in a character array for later use, just as you would store the value 12 in an int variable. The string library string.h provides several functions for manipulation strings. In this week’s lab, you will see static and dynamic ways of storing strings, and write your own versions of the string library function strcpy(), which copies one string into another.
Engineering
1 answer:
salantis [7]3 years ago
3 0

#include<stdio.h>

#include<string.h>

 

int main() {

  char str1[100];

  char str2[100];

 

  printf("\nEnter the String 1 : ");

  gets(str1);

 

  strcpy(str2, str1);

  printf("\nCopied String : %s", str2);

 

  return (0);

}

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1. Use the charges to create an electric dipole with a horizontal axis by placing a positive and a negative charge (equal in mag
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2)

a)  to the right of the dipole    E_total = kq [1 / (r + a)² - 1 / r²]

b)To the left of the dipole      E_total = - k q [1 / r² - 1 / (r + a)²]

c) at a point between the dipole, that is -a <x <a  

      E_total = kq [1 / x² + 1 / (2a-x)²]

d) on the vertical line at the midpoint of the dipole (x = 0)

E_toal = 2 kq 1 / (a ​​+ y)² cos θ

Explanation:

2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.

This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point

Let's analyze each point separately.

The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.

a) to the right of the dipole

The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity

           E₊ = k q / (r + a)²

where 2a is the distance between the charges of the dipole and the field is to the right

the negative charge creates an incoming field of magnitude

           E₋ = -k q / r²

The field is to the left

therefore the total field is the sum of these two fields

           E_total = E₊ + E₋

           E_total = kq [1 / (r + a)² - 1 / r²]

we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.

b) To the left of the dipole

The result is similar to the previous one by the opposite sign, since the closest charge is the positive one

E₊ is to the left and E₋ is to the right

          E_total = - k q [1 / r² - 1 / (r + a)²]

We see that this field is also directed to the left

c) at a point between the dipole, that is -a <x <a

In this case the E₊ field points to the right and the E₋ field points to the right

                      E₊ = k q 1 / x²

                      E₋ = k q 1 / (2a-x)²

                      E_total = kq [1 / x² + 1 / (2a-x)²]

in this case the field points to the right

d) on the vertical line at the midpoint of the dipole (x = 0)

    In this case the E₊ field points in the direction of the positive charge and the test charge

    in E₋ field the ni is between the test charge and the negative charge,

the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)

                      E₊ = kq 1 / (a ​​+ y) 2

                      E₋ = kp 1 / (a ​​+ y) 2

                      E_total = E₊ₓ + E_{-x}

                      E_toal = 2 kq 1 / (a ​​+ y)² cos θ

e) same as the previous part, but on the negative side

                        E_toal = 2 kq 1 / (a ​​+ y)² cos θ

When analyzing the previous answer there is no point where the field is zero

The different configurations are outlined in the attached

3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive

a) to the right

in this case the two field goes to the right

           E_total = kq [1 / (r + a)² + 1 / r²]

b) to the left

            E_total = - kq [1 / (r + a)² + 1 / r²]

c) between the two charges

E₊ goes to the right

E₋ goes to the left

            E_total = kq [1 / x² - 1 / (2a-x)²]

d) between vertical line at x = 0

             

E₊ salient between test charge and positive charge

           E_total = 2 kq 1 / (a ​​+ y)² sin θ

In this configuration at the point between the two charges the field is zero

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Answer: 7.500 2.99

Explanation:

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