Due to +q charge flux
"\\phi_1=\\frac{q}{\\epsilon_0}"
Due to +4q charge flux
"\\phi_1=\\frac{4q}{\\epsilon_0}"
Net flux
"\\phi=\\phi_1+\\phi_2"
"\\phi=\\frac{q}{\\epsilon_0}+\\frac{4q}{\\epsilon_0}=\\frac{5q}{\\epsilon_0}"
Electric field
"E_1=\\frac{kQ}{R^2}\\\\E_2=\\frac{4kq}{R^2}"
"E=E_2-E_1"
"E=\\frac{4kq}{R^2}-\\frac{kq}{R^2}=\\frac{3kq}{R^2}"
Net direction (E.F)of electric field E2 direction
Comments
Leave a comment