Deux charges ponctuelles q_(A)=-1mu Cetq_(B)=-4mu C sont placées en deux points A et B distants de 18 cm. 1) Montrer qu'il existe sur la droite (AB) un point M où le champ électrostatique résultant est nul.

The electric field created by a point charge *q* at a distance *r* is given by E = k*q/r², where k is Coulomb's constant. The electric field is a vector, directed away from positive charges and towards negative charges.<br /><br />For the resultant electric field at point M to be zero, the electric fields from charges q_A and q_B must be equal in magnitude and opposite in direction. Since both charges are negative, point M must lie between A and B. Let's assume the distance from A to M is 'x' cm. Then the distance from B to M is (18 - x) cm.<br /><br />The electric field due to q_A at M is E_A = k*|q_A|/x² towards A.<br />The electric field due to q_B at M is E_B = k*|q_B|/(18-x)² towards B.<br /><br />For the net field to be zero, E_A = E_B:<br /><br />k*|q_A|/x² = k*|q_B|/(18-x)²<br /><br />Since k is a constant, we can cancel it out:<br /><br />|q_A|/x² = |q_B|/(18-x)²<br /><br />Substituting the values of the charges:<br /><br />1/x² = 4/(18-x)²<br /><br />Taking the square root of both sides (since x and (18-x) are positive):<br /><br />1/x = 2/(18-x)<br /><br />Cross-multiplying:<br /><br />18 - x = 2x<br /><br />3x = 18<br /><br />x = 6 cm<br /><br />Therefore, there exists a point M between A and B, 6 cm away from A (and 12 cm away from B), where the resultant electric field is zero. This demonstrates that such a point exists.<br />