Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
Answer:
#include <iostream>
#include <vector>
using namespace std;
int main() {
const int NUM_GUESSES = 3;
vector<int> userGuesses(NUM_GUESSES);
int i = 0;
int uGuess = 0;
for(i = 0; i <= userGuesses.size() - 1; i++){
cin >> uGuess;
userGuesses.at(i) = uGuess;
}
cout << endl;
return 0;
}
Explanation:
First inbuilt library were imported. Then inside the main( ) function, 3 was assigned to NUM_GUESSES meaning the user is to guess 3 numbers. Next, a vector was defined with a size of NUM_GUESSES.
Then a for-loop is use to receive user guess via cin and each guess is assigned to the vector.
